In Mathematica, Residue[1/KleinInvariantJ[t], {t,Exp[2*pi*I/3]}]
results in a very ungainly expression involving complete elliptic integrals of the first and second kind evaluated at the modular lambda function $\lambda(\mathrm{e}^{2\pi i /3})$.
Is there a more elegant representation of this residue? (poking around for the numerical value on the Inverse Symbolic Calculator didn't return anything)
What about the residues of the other poles of $\frac{1}{J(\tau)}$?
I suppose the independent variable is the period ratio $\tau$.
I will express $1/J(\tau)$ as a Laurent series in $z = 2\pi\mathrm{i}(\tau - \tau_0)$ where $\tau_0$ is a zero of $J$. Using $z$ gives rise to the differential operator $$(\ )' = \frac{\mathrm{d}}{\mathrm{d}z} = \frac{1}{2\pi\mathrm{i}}\frac{\mathrm{d}}{\mathrm{d}\tau} = q\frac{\mathrm{d}}{\mathrm{d}q} \quad\text{where}\quad q = \exp(2\pi\mathrm{i}\tau)$$
I will express everything using Eisenstein series $\operatorname{E}_2,\operatorname{E}_4,\operatorname{E}_6$. These are holomorphic on all of $\mathbb{H}\cup\{\mathrm{i}\infty\}$ and are coupled by the following system of differential equations: $$\begin{align} \operatorname{E}_2' &= \frac{1}{12}\left(\operatorname{E}_2^2 - \operatorname{E}_4\right) \\\operatorname{E}_4' &= \frac{1}{3}\left(\operatorname{E}_2\operatorname{E}_4 - \operatorname{E}_6\right) \\\operatorname{E}_6' &= \frac{1}{2}\left(\operatorname{E}_2\operatorname{E}_6 - \operatorname{E}_4^2\right) \end{align}$$ Since $\operatorname{E}_2,\operatorname{E}_4,\operatorname{E}_6$ have no pairwise common zeros, the above system of equations implies that their derivatives are nonzero at their zeros. This means that all their zeros are simple.
Mathematica's KleinInvariantJ is
$$J = \frac{j}{1728}
= \frac{\operatorname{E}_4^3}{\operatorname{E}_4^3-\operatorname{E}_6^2}$$
The denominator is a multiple of the modular discriminant and therefore
finite and nonzero on $\mathbb{H}$.
The zeros of $J$ are therefore the zeros of $\operatorname{E}_4$
with three-fold multiplicity.
One of those zeros is located at
$\tau_0 = \zeta_3 = \exp(2\pi\mathrm{i}/3)$, but initially I will only assume
$\operatorname{E}_4(\tau_0) = 0$ for greater generality.
For the upcoming calculations, I will use a symbolic calculator,
in this case Pari/GP. GP has a marvellous diffop function
that allows you to specify a list of symbols and a list of their
derivatives, and it uses those to differentiate some symbolic expression.
This makes it easy to get a truncated Laurent series of $1/J$
around $z = 0$:
z; /* first symbol is treated as the main series variable */
sym = [E2, E4, E6];
dsym = [(E2^2 - E4)/12, (E2*E4 - E6)/3, (E2*E6 - E4^2)/2];
taylorcoeff(f,n) = subst(diffop(f, sym, dsym, n)/n!, E4, 0);
/* After having carried out the differentiation,
the coeffs are to be taken as values at z = 0 where E4 = 0 */
J = E4^3 / (E4^3 - E6^2);
Jser = sum(n = 3, 5, taylorcoeff(J, n) * z^n) + O(z^6);
/* We could have started with n = 0, but we already know
that the coeffs for n < 3 are zero */
/* Now let's ask for the series of 1/J in z */
1/Jser
This results in $$\frac{1}{J} = \left.\frac{27}{\operatorname{E}_6}\right|_{z=0} z^{-3} - \left.\frac{27\operatorname{E}_2}{4\operatorname{E}_6}\right|_{z=0} z^{-2} + \left.\frac{9\operatorname{E}_2^2}{16\operatorname{E}_6}\right|_{z=0} z^{-1} + \operatorname{O}(z^0)$$ The coefficient of $z^{-1}$ is the residue of $1/J$ at $z=0$ with respect to $z$. The residue that we actually want is the coefficient of $(\tau - \tau_0)^{-1} = 2\pi\mathrm{i} z^{-1}$. Therefore $$\operatorname{res}\left(\frac{1}{J}, \tau_0\right) = \frac{1}{2\pi\mathrm{i}} \frac{9\operatorname{E}_2(\tau_0)^2}{16\operatorname{E}_6(\tau_0)}$$
As shown in the answer to another question, the set of zeros of $J$ is the orbit of $\zeta_6$, or equivalently of $\zeta_3$, under the action of $\operatorname{SL}(2,\mathbb{Z})$. Therefore, let us assume that $$\tau_0 = \frac{a\zeta_3 + b}{c\zeta_3 + d} \quad\text{with}\quad \begin{pmatrix}a&b\\c&d\end{pmatrix}\in\operatorname{SL}(2,\mathbb{Z})$$ Applying the transformation identities $$\begin{align} \operatorname{E}_2\left(\frac{a\tau + b}{c\tau + d}\right) &= \frac{6c(c\tau+d)}{\pi\mathrm{i}} + (c\tau+d)^2\operatorname{E}_2(\tau) \\\operatorname{E}_6\left(\frac{a\tau + b}{c\tau + d}\right) &= (c\tau+d)^6\operatorname{E}_6(\tau) \end{align}$$ yields $$\operatorname{res}\left(\frac{1}{J}, \frac{a\zeta_3 + b}{c\zeta_3 + d}\right) = \frac{1}{2\pi\mathrm{i}} \frac{9\left(\frac{6c(c\zeta_3+d)}{\pi\mathrm{i}} + (c\zeta_3+d)^2\operatorname{E}_2(\zeta_3)\right)^2} {16\,(c\zeta_3+d)^6\operatorname{E}_6(\zeta_3)}$$ So all that remains is to find the values of $\operatorname{E}_2$ and $\operatorname{E}_6$ at $\zeta_3$. For $\operatorname{E}_2$ this is easy, just use the transformation properties: $$\begin{align} -\zeta_3^{-1} &= \zeta_3 + 1 \\\therefore\quad \operatorname{E}_2(\zeta_3) &= \operatorname{E}_2(\zeta_3 + 1) = \operatorname{E}_2(-\zeta_3^{-1}) = \frac{6\zeta_3}{\pi\mathrm{i}} + \zeta_3^2\operatorname{E}_2(\zeta_3) \\\therefore\quad \operatorname{E}_2(\zeta_3) &= \frac{6\zeta_3}{\pi\mathrm{i}(1 - \zeta_3^2)} = \frac{2\sqrt{3}}{\pi} \end{align}$$
For $\operatorname{E}_6$, we do not gain additional information from the transformation trick. Instead I'll use the following formula: $$\sqrt[6]{\operatorname{E}_6} = {}_2\mathrm{F}_1\left(\frac{1}{12},\frac{7}{12};1;\frac{1}{1 - J}\right)$$ which holds for every $\tau$ that can be reached from $\mathrm{i}\infty$ through a path where the hypergeometric series for the above ${}_2\mathrm{F}_1$ converges, which implies that $|1/(1 - J)|$ must not exceed $1$ anywhere on that path. Fortunately, this just barely holds for $\tau=\zeta_3$, and thus we get $$\sqrt[6]{\operatorname{E}_6(\zeta_3)} = {}_2\mathrm{F}_1\left(\frac{1}{12},\frac{7}{12};1;1\right)$$ Using the evaluation, reflection and multiplication formulae $$\begin{align} {}_2\mathrm{F}_1(u,v;w;1) &= \frac{\Gamma(w)\,\Gamma(w - u - v)}{\Gamma(w - u)\,\Gamma(w - v)} \quad\text{for}\quad\{w, w - u - v\}\cap\{0,-1,-2,\ldots\} = \emptyset \\\Gamma(s)\,\Gamma(1-s) &= \frac{\pi}{\sin(\pi s)} \quad\text{for}\quad s\in\mathbb{C}\setminus\mathbb{Z} \\\Gamma(ms) &= m^{ms-\frac{1}{2}} \frac{\Gamma(s)\,\Gamma\left(s+\frac{1}{m}\right)\cdots \Gamma\left(s+\frac{m-1}{m}\right)}{(2\pi)^{\frac{m-1}{2}}} \quad\text{for}\quad m\in\mathbb{N} \end{align}$$ we get $$\begin{align} \sqrt[6]{\operatorname{E}_6(\zeta_3)} &= \frac{\Gamma\left(\frac{1}{3}\right)} {\Gamma\left(\frac{5}{12}\right)\Gamma\left(\frac{11}{12}\right)} = \frac{\sqrt{3}\,\Gamma^3\left(\frac{1}{3}\right)}{2^{3/2}\pi^2} = \frac{2^{3/2}\pi}{3\,\Gamma^3\left(\frac{2}{3}\right)} \\\therefore\quad \operatorname{E}_6(\zeta_3) &= \frac{3^3\,\Gamma^{18}\left(\frac{1}{3}\right)}{2^9\pi^{12}} = \frac{2^9\pi^6}{3^6\,\Gamma^{18}\left(\frac{2}{3}\right)} \end{align}$$ Plugging those values for $\operatorname{E}_2$ and $\operatorname{E}_6$ into the residue formula, we finally get $$\operatorname{res}\left(\frac{1}{J}, \zeta_3\right) = -\frac{3^9\,\Gamma^{18}\left(\frac{2}{3}\right)}{2^{12}\pi^9}\mathrm{i}$$ This should match your numeric result. (I have not tested it.) I leave it to you to polish the residue formula for general $\tau_0$.
