I will slightly alter your notation. Let $$ J(\Theta)=\frac{-1}{m}\sum_{i=1}^{m}\sum_k y_k^{(i)}\log(\hat{p}_k^{(i)}),\;\;\; \hat{p}_k^{(i)}=\frac{\exp(\theta_k^Tx^{(i)})}{\sum_\alpha \exp(\theta_\alpha^Tx^{(i)})}$$
where $\theta_\xi\in\mathbb{R}^n$, $\Theta\in\mathbb{R}^{K \times n}$.
Since $J$ is a scalar function, the derivative $\nabla_\Theta J(\Theta)$ must be a matrix. We will compute component $j,\ell$ of that matrix.
$$ \log(\hat{p}_k^{(i)}) = \theta_k^Tx^{(i)} - \log\sum_\alpha\exp(\theta_\alpha^Tx^{(i)}) $$
$$ \nabla_{\theta_{j\ell}} J(\Theta)= \frac{-1}{m}\sum_{i=1}^{m}\sum_k y_k^{(i)}\partial_{j\ell}\log(\hat{p}_k^{(i)}) $$
where $\nabla_{\theta_{j\ell}}=\partial/\partial \theta_{j\ell}=\partial_{j\ell}$. The first term:
$$ \partial_{j\ell}\, \theta_k^Tx^{(i)} = \sum_\beta \partial_{j\ell} \theta_{k\beta}x^{(i)}_\beta = \delta_{kj} x_\ell^{(i)} $$
where $\delta_{\xi\eta}$ is the Kronecker Delta.
Define: $$ S_i(\Theta)=\sum_\alpha \exp(\theta_\alpha^Tx^{(i)}) $$
Now the second term:
\begin{align}
\partial_{j\ell} \log\sum_\alpha\exp(\theta_\alpha^Tx^{(i)})
&= S_i(\Theta)^{-1}\partial_{j\ell} \sum_\alpha \exp(\theta_\alpha^Tx^{(i)})\\
&= S_i(\Theta)^{-1} \sum_\alpha \exp(\theta_\alpha^Tx^{(i)}) \partial_{j\ell}[\theta_\alpha^Tx^{(i)}]\\
&= S_i(\Theta)^{-1} \sum_\alpha \exp(\theta_\alpha^Tx^{(i)}) \delta_{\alpha j} x^{(i)}_\ell \\ &= S_i(\Theta)^{-1} \exp(\theta_j^Tx^{(i)}) x^{(i)}_\ell \\
&= x^{(i)}_\ell \hat{p}_j^{(i)}
\end{align}
Putting it all together gives me: $$ \nabla_{\theta_{j\ell}} J(\Theta) = \frac{-1}{m}\sum_{i=1}^m\sum_k y_k^{(i)} \left[ \delta_{kj} x_\ell^{(i)} - x^{(i)}_\ell \hat{p}_j^{(i)} \right]=\frac{1}{m}\sum_{i}x^{(i)}_\ell\sum_k y_k^{(i)} \left[ \hat{p}_j^{(i)} - \delta_{kj} \right] $$
We can simplify the inner sum by continuing this index masochism. We will show that:
$$ \sum_k y_k^{(i)} \left[ \hat{p}_j^{(i)} - \delta_{kj} \right] = y^{(i)}_j\left[ \hat{p}_j^{(i)} - 1 \right] + \sum_{k\ne j} \hat{p}_j^{(i)}y^{(i)}_k = \hat{p}_j^{(i)} - y^{(i)}_j $$
There are two cases for $\partial_{j\ell} J(\Theta)$, depending on the value of $y^{(i)}_j$.
\begin{align}
&\text{Case 1: } y_j^{(i)}=1 \;\;\;\implies\;\;\;
y^{(i)}_j\left[ \hat{p}_j^{(i)} - 1 \right] + \sum_{k\ne j} \hat{p}_j^{(i)}\underbrace{y^{(i)}_k}_0
= \underbrace{y^{(i)}_j}_1 \hat{p}_j^{(i)} - y^{(i)}_j
= \hat{p}_j^{(i)} - y^{(i)}_j \\
&\text{Case 2: } y_p^{(i)}=1,\,p\ne j \;\;\;\implies\;\;\;
\underbrace{y^{(i)}_j}_0 \left[ \hat{p}_j^{(i)} - 1 \right] + \sum_{k\ne j} \hat{p}_j^{(i)}\underbrace{y^{(i)}_k}_{\delta_{pk}}
= \hat{p}_j^{(i)} - \underbrace{y^{(i)}_j}_0 \\
\therefore \;\; & \;\; \sum_k y_k^{(i)} \left[ \hat{p}_j^{(i)} - \delta_{kj} \right] = \hat{p}_j^{(i)} - {y^{(i)}_j}
\end{align}
Substituting this into our expression gives:
$$ \nabla_{\theta_{j\ell}} J(\Theta) = \frac{-1}{m}\sum_{i=1}^m\sum_k y_k^{(i)} \left[ \delta_{kj} x_\ell^{(i)} - x^{(i)}_\ell \hat{p}_j^{(i)} \right]=\frac{1}{m}\sum_{i}x^{(i)}_\ell \left[ \hat{p}_j^{(i)} - {y^{(i)}_j} \right] $$
Finally, we can gather together the parts of the Jacobian corresponding to the $j$th row, i.e. $\theta_j$, which is a vector $\nabla_{\theta_j} J(\Theta)\in\mathbb{R}^{n}$. This is equivalent to simply grouping together the $\ell$ indices into the vector:
$$ \nabla_{\theta_{j}} J(\Theta) =\frac{1}{m}\sum_{i}x^{(i)} \left[ \hat{p}_j^{(i)} - {y^{(i)}_j} \right] $$
Relevant links: [1], [2], [3], [4], [5], [6], [7], [8], [9], [10], [11]
