Unfortunately we cannot directly use the reciprocal identity because of a subtle distinction between the derivatives. $\frac{\partial x}{\partial \zeta}$ is done holding $\eta$ constant and $\frac{\partial \zeta}{\partial x}$ holds $y$ constant, but in the first one varying $\zeta$ leads to a variation in $y$, so they cannot be measuring the same thing.
To derive a rule, take the equation $\zeta = \zeta(x(\zeta,\eta),y(\zeta,\eta))$ and differentiate it on both sides:
$$\frac{\partial \zeta}{\partial \zeta} = 1 = \frac{\partial \zeta}{\partial x}\frac{\partial x}{\partial \zeta} + \frac{\partial \zeta}{\partial y}\frac{\partial y}{\partial \zeta}$$
$$\frac{\partial \zeta}{\partial \eta} = 0 = \frac{\partial \zeta}{\partial x}\frac{\partial x}{\partial \eta} + \frac{\partial \zeta}{\partial y}\frac{\partial y}{\partial \eta}$$
Use the second equation to solve for $\frac{\partial \zeta}{\partial y}$ then plug in and rearrange the first to solve for $\frac{\partial \zeta }{\partial x}$. We get the following equations:
$$\frac{\partial \zeta}{\partial x} = \frac{\frac{\partial y}{\partial \eta}}{\frac{\partial x}{\partial \zeta}\frac{\partial y}{\partial \eta}-\frac{\partial x}{\partial \eta}\frac{\partial y}{\partial \zeta}}$$
$$\frac{\partial \zeta}{\partial y} = \frac{-\frac{\partial x}{\partial \eta}}{\frac{\partial x}{\partial \zeta}\frac{\partial y}{\partial \eta}-\frac{\partial x}{\partial \eta}\frac{\partial y}{\partial \zeta}}$$
where we recognize the denominator as the (signed) determinant of the Jacobian matrix
$$J=\begin{pmatrix} \frac{\partial x}{\partial \zeta} & \frac{\partial x}{\partial \eta} \\ \frac{\partial y}{\partial \zeta} & \frac{\partial y}{\partial \eta} \\ \end{pmatrix}$$
To get the second derivatives, differentiate the above system again:
$$\frac{\partial^2 \zeta}{\partial \zeta^2} = 0 = \frac{\partial^2 \zeta}{\partial x^2}\left(\frac{\partial x}{\partial \zeta}\right)^2 + 2\frac{\partial^2 \zeta}{\partial x \partial y}\frac{\partial x}{\partial \zeta}\frac{\partial y}{\partial \zeta} + \frac{\partial^2 \zeta}{\partial y^2}\left(\frac{\partial y}{\partial \zeta}\right)^2 + \frac{\partial \zeta}{\partial x}\frac{\partial^2 x}{\partial \zeta^2} + \frac{\partial \zeta}{\partial y}\frac{\partial^2 y}{\partial \zeta^2}$$
$$\frac{\partial^2 \zeta}{\partial \eta^2} = 0 = \frac{\partial^2 \zeta}{\partial x^2}\left(\frac{\partial x}{\partial \eta}\right)^2 + 2\frac{\partial^2 \zeta}{\partial x \partial y}\frac{\partial x}{\partial \eta}\frac{\partial y}{\partial \eta} + \frac{\partial^2 \zeta}{\partial y^2}\left(\frac{\partial y}{\partial \eta}\right)^2 + \frac{\partial \zeta}{\partial x}\frac{\partial^2 x}{\partial \eta^2} + \frac{\partial \zeta}{\partial y}\frac{\partial^2 y}{\partial \eta^2}$$
$$\frac{\partial^2 \zeta}{\partial \eta \partial \zeta} = 0 = \frac{\partial^2 \zeta}{\partial x^2}\frac{\partial x}{\partial \eta}\frac{\partial x}{\partial \zeta} + \frac{\partial^2 \zeta}{\partial x \partial y}\left(\frac{\partial x}{\partial \zeta}\frac{\partial y}{\partial \eta}+\frac{\partial x}{\partial \eta}\frac{\partial y}{\partial \zeta}\right) + \frac{\partial^2 \zeta}{\partial y^2}\frac{\partial y}{\partial \eta}\frac{\partial y}{\partial \zeta} + \frac{\partial \zeta}{\partial x}\frac{\partial^2 x}{\partial \eta \partial \zeta} + \frac{\partial \zeta}{\partial y}\frac{\partial^2 y}{\partial \eta \partial \zeta}$$
We have a system of three equations for the three second derivatives. Since the equations have become unwieldy, we will switch to the subscript notation for partial derivatives. Use the third equation to substitute for the mixed derivative in the other two equations and substitute in the first derivatives from earlier:
$$\zeta_{xy} = -\frac{\zeta_{xx}x_\eta x_\zeta + \zeta_{yy}y_\eta y_\zeta + \zeta_x x_{\zeta \eta} + \zeta_y y_{\zeta \eta}}{x_\zeta y_\eta + x_\eta y_\zeta}$$
$$\implies \begin{cases} 0 = \zeta_{xx}x_\zeta^2(x_\zeta y_\eta - x_\eta y_\zeta)^2-\zeta_{yy}y_\zeta^2(x_\zeta y_\eta - x_\eta y_\zeta)^2 + (x_{\zeta \zeta}y_\eta - y_{\zeta \zeta}x_\eta)(x_\zeta y_\eta + x_\eta y_\zeta) -2 x_\zeta y_\zeta(x_{\zeta \eta} + y_{\zeta \eta}) \\ 0 = -\zeta_{xx}x_\eta^2(x_\zeta y_\eta - x_\eta y_\zeta)^2 + \zeta_{yy}y_\eta^2(x_\zeta y_\eta - x_\eta y_\zeta)^2 + (x_{\eta \eta}y_\eta - y_{\eta \eta}x_\eta)(x_\zeta y_\eta + x_\eta y_\zeta) -2 x_\eta y_\eta(x_{\zeta \eta} + y_{\zeta \eta})\\ \end{cases}$$
$$ \equiv \begin{cases} 0 = \zeta_{xx}x_\zeta^2(x_\zeta y_\eta - x_\eta y_\zeta)^2-\zeta_{yy}y_\zeta^2(x_\zeta y_\eta - x_\eta y_\zeta)^2 + G \\ 0 = -\zeta_{xx}x_\eta^2(x_\zeta y_\eta - x_\eta y_\zeta)^2 + \zeta_{yy}y_\eta^2(x_\zeta y_\eta - x_\eta y_\zeta)^2 + H \\ \end{cases}$$
to shorthand the terms on the far right because they are completely in terms of $\zeta$ and $\eta$ derivatives of $x$ and $y$.
This leads us to our final answers of
$$\frac{\partial^2 \zeta}{\partial x^2} = \frac{-(y_\zeta^2 H + y_\eta^2 G)}{(x_\zeta y_\eta - x_\eta y_\zeta)^2(x_\zeta^2 y_\eta^2 - x_\eta^2 y_\zeta^2)} = \frac{-(y_\zeta^2 H + y_\eta^2 G)}{(x_\zeta y_\eta - x_\eta y_\zeta)^3(x_\zeta y_\eta + x_\eta y_\zeta)}$$
$$\frac{\partial^2 \zeta}{\partial y^2} = \frac{-(x_\zeta^2 H + x_\eta^2 G)}{(x_\zeta y_\eta - x_\eta y_\zeta)^2(x_\zeta^2 y_\eta^2 - x_\eta^2 y_\zeta^2)} = \frac{-(x_\zeta^2 H + x_\eta^2 G)}{(x_\zeta y_\eta - x_\eta y_\zeta)^3(x_\zeta y_\eta + x_\eta y_\zeta)}$$
The same procedure can be done for $\eta$ (since the answer should be symmetric, just take this answer and swap all $\zeta \leftrightarrow \eta$ ), but the lesson here is never do multidimensional chain rule on more than a first derivative ever again.
