Proving: $\int_0^1 \int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy=\frac{225}{2}\zeta(5)$ 
Proving:$$\displaystyle\int_0^1\displaystyle\int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy=\frac{225}{2}\zeta(5)$$

I tried using variable switching
$\ln(xy)=t$  But I did not reach any results after the calculation
\begin{align*}
k&=\displaystyle\int_0^1\displaystyle\int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dx\\
&=\displaystyle\int_0^1\displaystyle\int_{-\infty}^{\ln(y)}\frac{t^4e^t}{(1+e^t)^2y^2}dtdy\\
&=\displaystyle\int_0^1\displaystyle\int_{-\infty}^{\ln(y)}\frac{t^4e^t}{y^2(1+e^t)}\displaystyle\sum_{n=0}^{\infty}(-e^t)dtdy\\
&=\displaystyle\int_0^1\frac{1}{y^2}\left(\displaystyle\sum_{n=0}^{\infty}\displaystyle\int_{-\infty}^{\ln(y)}\frac{t^4(-e^{2t})}{1+e^t}dt\right)dy\\
&=\displaystyle\int_0^1\frac{1}{y^2}\left(\displaystyle\sum_{n=0}^{\infty}\displaystyle\int_{\ln(y)}^{\infty}\frac{t^4e^{2t}}{1+e^t}dt\right)dy\\
\end{align*}

Starting off after your first substitution, notice that $$\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}, \text{ for } x \in(-1,1)$$
Since the domain of $x,y$ is $(0,1)$, we can write $$\frac{e^t}{(1+e^t)^2}=-\sum_{n=1}^{\infty} {(-1)}^n n e^{tn}$$
In addition, you made a slight error when calculating $dt$ I suppose.  It should be $y$ not $y^2$ in the denominator.
$$\int_0^1 \int_0^{\ln{y}} \frac{t^4}{y}\left( -\sum_{n=1}^{\infty} {(-1)}^n n e^{tn}\right) \; dt \; dy$$
Because the summation converges, we can interchange the summation and integral sign from Fubini's theorem:
\begin{align}
k &= -\sum_{n=1}^{\infty} {(-1)}^n n \int_0^1 \frac{1}{y} \int_0^{\ln{y}} t^4 e^{tn}\; dt \; dy \\
&\overset{\text{IBP}}= -\sum_{n=1}^{\infty} {(-1)}^n n \int_0^1 \frac{y^{n-1} \left(n^4  \ln^4{y}-4n^3\ln^3{y}+12n^2\ln^2{y}-24n\ln{y}+25\right)}{n^5} \; dy \\
&\overset{\text{IBP}}= -\sum_{n=1}^{\infty} {(-1)}^n n \cdot \frac{120}{n^6} \\
&= 120\sum_{n=1}^{\infty} \frac{{(-1)}^{n+1}}{n^5} \\
&= \boxed{\frac{225}{2}\zeta(5)}  \\
\end{align}
