I am asked to prove that $$\frac{d}{dt}(\Pi(t)) = \Sigma(t)\Pi(t)$$ and deduce that $$np(n) = \sum_{k = 1}^n \sigma(n)p(n - k).$$ Here $\Pi(t)$ is the generating function for the partition function $p(n)$. Further, $\Sigma(t)$ is the generating function for the divisor function $\sigma(n)$, however this generating function is defined as follows $$\Sigma(t) = \sum_{n \geq 1}\sigma(n)t^{n - 1}$$.
Now, so far, I have gotten to this point
\begin{equation*} \begin{split} \frac{d}{dt}\left(\Pi(t)\right) &= \frac{d}{dt}\left(\prod_{n = 1}^\infty \frac{1}{1 - t^n}\right)\\ &= \sum_{n = 1}^\infty \frac{nt^{n - 1}}{(1 - t^n)^2}\left(\prod_{n = 1}^\infty \frac{1}{1 - t^n}\right)\\ \end{split} \end{equation*}
but I am not sure what to do with left hand side. Any help would be greatly appreciated.
There is a slight error in your derivative computation. We have $$\frac{d}{dt}(\Pi(t)) = \sum_{n=1}^{\infty}\frac{nt^{n - 1}}{(1 - t^n)^2}\left( \prod_{\substack{m \geq 1 \\ m \neq n}} \frac{1}{1 - t^m}\right) = \sum_{n=1}^{\infty} \frac{nt^{n - 1}}{(1 - t^n)}\prod_{m=1}^{\infty} \frac{1}{1 - t^m} = \sum_{n=1}^{\infty} \frac{nt^{n - 1}}{(1 - t^n)} \cdot \Pi(t).$$
Now expand $\frac{1}{1 - t^n}$ as a geometric series to obtain $$\sum_{n=1}^{\infty} \frac{nt^{n-1}}{(1 - t^n)} = \sum_{n=1}^{\infty} nt^{n - 1}\left(\sum_{m=0}^{\infty}t^{nm}\right) = \sum_{\substack{n \geq 1 \\ m \geq 0}}nt^{n(m + 1) - 1} = \sum_{n,m \geq 1}nt^{nm - 1}.$$ Collecting together like powers of $t$, we see that $$\sum_{n,m \geq 1}nt^{nm - 1} = \sum_{n=1}^{\infty}\sigma(n)t^{n - 1}$$
Then $$\frac{d}{dt}(\Pi(t)) = \sum_{n=1}^{\infty}\sigma(n)t^{n - 1}\sum_{m=0}^{\infty}p(m)t^m = \sum_{\substack{n \geq 1 \\ m \geq 0}}\sigma(n)p(m)t^{n + m - 1}.$$ Collecting like powers of $t$, the right-hand side becomes $$\sum_{n=1}^{\infty}\left(\sum_{k=1}^n \sigma(n)p(n - k)\right) t^{n - 1}$$ Since we also have $$\frac{d}{dt}(\Pi(t)) = \sum_{n = 1}^{\infty}np(n)t^{n - 1},$$ comparing coefficients shows that $$np(n) = \sum_{k=1}^n \sigma(n)p(n - k)$$