Prove $\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\sum_{j=1}^{2k}\frac{\left ( -1 \right )^{j}}{j}=\frac{\pi ^{2}}{48}+\frac{1}{4}\ln^22$.

Yes! As you mentioned, it's a very useful formula.

Use $$\sum_{j=1}^{n}\frac{\left ( -1 \right )^{j-1}}{j}=\ln 2+\left ( -1 \right )^{n-1}\int_{0}^{1}\frac{x^{n}}{1+x}\, \mathrm{d}x$$ we get \begin{align*} &\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\sum_{j=1}^{2k}\frac{\left ( -1 \right )^{j}}{j}=\ln 2\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}-\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\int_{0}^{1}\frac{x^{2k}}{1+x}\, \mathrm{d}x\\ &=\ln^22+\int_{0}^{1}\frac{1}{1+x}\sum_{k=1}^{\infty }\frac{\left ( -x^{2} \right )^{k}}{k}\, \mathrm{d}x=\ln^22-\int_{0}^{1}\frac{\ln\left ( 1+x^{2} \right )}{1+x}\, \mathrm{d}x \end{align*} let $$f\left ( t \right )=\int_{0}^{1}\frac{\ln\left ( 1+tx^{2} \right )}{1+x}\,\mathrm{d}x$$ then \begin{align*} f{}'\left ( t \right ) &=\int_{0}^{1}\frac{x^{2}}{\left ( 1+x \right )\left ( 1+tx^{2} \right )}\,\mathrm{d}x \\ &=\frac{1}{t+1}\int_{0}^{1}\frac{x-1}{1+tx^{2}}\,\mathrm{d}x+\frac{1}{t+1}\int_{0}^{1}\frac{\mathrm{d}x}{x+1} \\ &=\frac{1}{t+1}\left [ \frac{1}{2t}\ln\left ( 1+tx^{2} \right )-\frac{1}{\sqrt{t}}\arctan\left ( \sqrt{t}x \right )+\ln\left ( x+1 \right ) \right ]_{0}^{1} \\ &=\frac{1}{t+1}\left [ \frac{1}{2t}\ln\left ( 1+t \right )-\frac{1}{\sqrt{t}}\arctan\left ( \sqrt{t} \right )+\ln 2 \right ] \end{align*} Integrate back $$\Rightarrow f\left ( t \right )=\frac{1}{2}\left [ -\mathrm{Li}_{2}\left ( -t \right )-\frac{1}{2}\ln^{2}\left ( t+1 \right ) \right ]-\arctan^{2}\sqrt{t}+\ln 2\ln\left ( t+1 \right )$$ then let $t=1$ and use $\displaystyle \mathrm{Li}_{2}\left ( -1 \right )=-\frac{\pi ^{2}}{12}$, you will get the answer as wanted.

EDIT: $$\mathrm{Li}_{2}\left ( -1 \right )=\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k^{2}}=\sum_{k=1}^{\infty }\frac{1}{k^{2}}-2\sum_{k=1}^{\infty }\frac{1}{\left ( 2n-1 \right )^{2}}=\frac{\pi ^{2}}{6}-2\cdot \frac{\pi ^{2}}{8}=-\frac{\pi ^{2}}{12}$$