prove $\sum_{n=1}^\infty \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$ this series was evaluated by Cornel Valean here using series manipulation. 
I took a different path as follows:
using the identity:$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\right)$$
multiply both sides by $\ln^3x/x$ then integrate
$$-6\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n^4}=\int_0^1\frac{\ln^2(1-x)\ln^3x}{x(1-x)}\ dx$$
I was able here to find
\begin{align}
\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}&=\frac43\zeta^2(3)-\frac23\sum_{k=1}^\infty\frac{H_k^{(3)}}{k^3}\\
&=\zeta^2(3)-\frac13\zeta(6)
\end{align}
as for the integral, it seems very tedious to calculate it using the derivative of beta function. 
can we find it with or without using beta function? 
 A: This solution is by Cornel Valean.
Using the follwing identity: ( see Lemma $2(b)$ in this paper)
 $$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$
and since $$\int_0^1x^{n-1}\ln^2(1-x)\ dx=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\int_0^1x^{n+k-1}\ dx=2\sum_{k=1}^\infty\frac{H_{k-1}}{k(n+k)}$$ 
Then $$\sum_{k=1}^\infty\frac{H_{k-1}}{k(n+k)}=\frac{H_n^2+H_n^{(2)}}{2n}\tag{1}$$
Divide both sides by $n^3$ then sum both sides from $n=1$ to $\infty$, we get
\begin{align}
S&=\color{blue}{\frac12\sum_{n=1}^\infty\frac{H_n^2}{n^4}+\frac12\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}}=\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\sum_{n=1}^\infty\frac{1}{n^3(n+k)}\right)\\
&=\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\sum_{n=1}^\infty\left[\frac{1}{k^3}\left(\frac{1}{n}-\frac{1}{n+k}\right)-\frac1{n^2k^2}+\frac1{n^3k}\right]\right)\\
&=\sum_{k=1}^\infty\left(\frac{H_k}{k}-\frac{1}{k^2}\right)\left(\frac{H_k}{k^3}-\frac{\zeta(2)}{k^2}+\frac{\zeta(3)}{k}\right)\\
&=\sum_{k=1}^\infty\frac{H_k^2}{k^4}-\sum_{k=1}^\infty\frac{H_k}{k^5}-\zeta(2)\sum_{k=1}^\infty\left(\frac{H_k}{k^3}-\frac1{k^4}\right)+\zeta(3)\sum_{k=1}^\infty\left(\frac{H_k}{k^2}-\frac1{k^3}\right)\\
&=\sum_{k=1}^\infty\frac{H_k^2}{k^4}-\left(\frac74\zeta(6)-\frac12\zeta^2(3)\right)-\zeta(2)\left(\frac14\zeta(4)\right)+\zeta(3)\left(\zeta(3)\right)\\
&=\color{blue}{\sum_{k=1}^\infty\frac{H_k^2}{k^4}-\frac{35}{16}\zeta(6)+\frac32\zeta^2(3)}
\end{align}
Rearranging the blue sides, we get 

$$\sum_{k=1}^\infty\frac{H_k^2}{k^4}=\frac{35}{8}\zeta(6)-3\zeta^2(3)+\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}\\
=\frac{97}{24}\zeta(6)-2\zeta^2(3)$$

where we used $\ \displaystyle\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}=\zeta^2(3)-\frac13\zeta(6)\ $ (can be found in the same paper I linked or here)
A: For a slight variation on a theme.
As 
$$\int_0^1 x^{n - 1} \ln^2 (1 - x) \, dx = \frac{H^2_n}{n} + \frac{H^{(2)}_n}{n},$$
for a proof of this result, see here, we can write the sum as
\begin{align}
\sum_{n = 1}^\infty \frac{H^2_n}{n^4} &= \sum_{n = 1}^\infty \frac{1}{n^3} \cdot \frac{H^2_n}{n}\\
&= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + \int_0^1 \frac{\ln^2 (1 - x)}{x} \sum_{n = 1}^\infty \frac{x^n}{n^3} \, dx\\
&= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + \int_0^1 \frac{\ln^2 (1 - x) \operatorname{Li}_3 (x)}{x} \, dx.\tag1
\end{align}
Making use of the following Maclaurin series expansion for $\ln^2 (1 - x)$, namely
$$\ln^2 (1 - x) = 2 \sum_{n = 1}^\infty \frac{H_n x^{n + 1}}{n + 1},$$
the integral in (1) can be re-written as
\begin{align}
\sum_{n = 1}^\infty \frac{H^2_n}{n^4} &= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \sum_{n = 1}^\infty \frac{H_n}{n + 1} \underbrace{\int_0^1 x^n \operatorname{Li}_3 (x) \, dx}_{\text{IBP 3 times}}\\
&= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \sum_{n = 1}^\infty \frac{H_n}{n+ 1} \left [\frac{\zeta (3)}{n + 1} - \frac{\zeta (2)}{(n + 1)^2} + \frac{H_{n + 1}}{(n + 1)^3} \right ]\\
&= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \zeta (3) \underbrace{\sum_{n = 1}^\infty \frac{H_n}{(n + 1)^2}}_{n \, \mapsto \, n - 1} -2 \zeta (2) \underbrace{\sum_{n = 1}^\infty \frac{H_n}{(n + 1)^3}}_{n \, \mapsto \, n - 1} + 2 \underbrace{\sum_{n = 1}^\infty \frac{H_n H_{n + 1}}{(n + 1)^2}}_{n \, \mapsto \, n - 1}\\
&= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \zeta (3) \sum_{n = 1}^\infty \frac{1}{n^2} \left (H_n - \frac{1}{n} \right ) - 2 \zeta (2) \sum_{n = 1}^\infty \frac{1}{n^3} \left (H_n - \frac{1}{n} \right )\\
& \qquad + 2 \sum_{n = 1}^\infty \frac{H_n}{n^4} \left (H_n - \frac{1}{n} \right )\\ 
&= - \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} + 2 \zeta (3) \sum_{n = 1}^\infty \frac{H_n}{n^2} - 2 \zeta^2 (3) - 2 \zeta (2) \sum_{n = 1}^\infty \frac{H_n}{n^3} + 2 \zeta (2) \zeta (4)\\
& \qquad + 2 \sum_{n = 1}^\infty \frac{H^2_n}{n^4} - 2 \sum_{n = 1}^\infty \frac{H_n}{n^5}\\
\Rightarrow \sum_{n = 1}^\infty \frac{H^2_n}{n^4} &= \sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} - 2 \zeta (3) \sum_{n = 1}^\infty \frac{H_n}{n^2} + 2 \zeta (2) \sum_{n = 1}^\infty \frac{H_n}{n^3} + 2 \sum_{n = 1}^\infty \frac{H_n}{n^5}\\
& \qquad + 2 \zeta^2 (3) - 2 \zeta (2) \zeta (4).\tag2
\end{align}
Making use of the following results:
\begin{align}
\sum_{n = 1}^\infty \frac{H_n}{n^2} &= 2 \zeta (3)\\
\sum_{n = 1}^\infty \frac{H_n}{n^3} &= \frac{5}{4} \zeta (4)\\
\sum_{n = 1}^\infty \frac{H_n}{n^5} &= -\frac{1}{2} \zeta^2 (3) + \frac{7}{4} \zeta (6)\\
\sum_{n = 1}^\infty \frac{H^{(2)}_n}{n^4} &= \zeta^2 (3) - \frac{1}{3} \zeta (6)\\
\zeta (2) \zeta (4) &= \frac{7}{6} \zeta (6)
\end{align}
substituting into (2) leads to
$$\sum_{n = 1}^\infty \frac{H^2_n}{n^4} = \frac{97}{24} \zeta (6) - 2 \zeta^2 (3),$$
as desired.
