Here is a very different proof :

Let $0 < x_1 < x_2 < x_3 < \cdots $ be the positive roots of equation

$$\tan(x)=x \tag 0$$

These roots are such that

$$\sin(x_k)= x_k \cos(x_k) \tag 1 $$

(this relationship will be used later on).

We have to show that :

$$\sum_{k=1}^\infty \frac{1}{x_k^2}=\frac{1}{10} \tag 2$$

The framework is Hilbert space $L^2[0,1]$ of square integrable functions on $[0,1]$ with bilinear form $\langle f|g\rangle:=\int_0^1fg$.

Let us consider its closed subspace defined by

$$E=\{f\mid f(0)=0 \ \ \& \ \ f(1)=f'(1) \ \ \& f \in C^2\}, \tag 3 $$

itself a Hilbert space.

Operator $\Delta : f \mapsto f''$ is a self-adjoint operator on $E$, due to the fact that a double integration by parts gives :

$$\int f g'' = \int f'' g \ \ \iff \ \ \langle f|\Delta g \rangle=\langle \Delta f| g \rangle \tag 4 $$

The kernel of $\Delta$ is one-dimensional : it is the set of linear functions $f(x)=ax$. Let $\varphi_0(x)=x$ be the basis of this kernel.

Functions defined by $\varphi_k(x)=\sin(x_kx)$ (with $x_k$ as defined in (2)), due to property (1), belong to space $E$ and moreover are such that :

$$\Delta \varphi_k = x_k^2 \varphi_k \tag 5 $$

i.e., they are eigenvectors of operator $\Delta$ with associated eigenvalues $x_k^2$.

Their norms are found to be

$$\|\varphi_0\|^2=\frac{1}{3} \ \ \text{and} \ \ \|\varphi_k\|^2=\frac{1}{2}\sin^2(x_k)\tag{6}$$

Being eigenvectors of a self-adjoint operator, functions $\varphi_k$ are mutually orthogonal. We admit that they constitute an Hilbert basis of $E$, allowing us to write, for any $f \in E$ :

$$f=\sum_{k=0}^\infty \frac{1}{\|\varphi_k\|^2} \langle f
\mid \varphi_k \rangle\tag{7}$$

Now consider the particular function $f(x)=x^3-2x^2 \in E$.

We have the following chain of equalities :

$$\langle f\mid \varphi_k \rangle=\langle f\mid \frac{1}{x_k^2}\Delta\varphi_k \rangle=\frac{1}{x_k^2}\langle \Delta f\mid\varphi_k \rangle=\frac{1}{x_k^2}(-6\langle \varphi_0 \mid \varphi_k \rangle+ \langle 4\mid\varphi_k \rangle) = \frac{4}{x_k^2}(1-\cos(x_k))\tag{8}$$

$$\langle f\mid \varphi_0 \rangle=\int_0^1 x(x^3-2x^2) \, dx=-\tfrac{3}{10}\tag{9}$$

Now, using (7)

$$f(x):=x^3-2x^2=-\frac{9}{10}x+\sum_{k=1}^\infty \frac{8}{x_k^2(1+\cos(x_k))}\sin(x_kx)\tag{10}$$

Let us terminate by computing, thanks to (10), the expression $f'(0)+f'(1)$ in two ways (with the LHS and the RHS), from which we deduce (2).