Let [tex]S[/tex] denote the sum. We can first resolve the sum in [tex]m[/tex] by factorizing and decomposing into partial fractions.
[tex]\displaystyle S = \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{(-1)^{n+1}}{mn^2 + mn + m^2n} \\\\ ~~~~ = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n \sum_{m=1}^\infty \frac1{m(m+n+1)} \\\\ ~~~~ = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)} \sum_{m=1}^\infty \left(\frac1m - \frac1{m+n+1}\right)[/tex]
Rewrite the [tex]m[/tex]-summand as a definite integral. Interchange the integral and sum, and evaluate the resulting geometric sums.
[tex]\displaystyle \sum_{m=1}^\infty \left(\frac1m - \frac1{m+n+1}\right) = \sum_{m=1}^\infty \int_0^1 \left(x^{m-1} - x^{m+n}\right) \, dx \\\\ ~~~~~~~~ = \int_0^1 \sum_{m=1}^\infty \left(x^{m-1} - x^{m+n}\right) \, dx \\\\ ~~~~~~~~ = \int_0^1 \frac{1 - x^{n+1}}{1 - x} \, dx \\\\ ~~~~~~~~ = \int_0^1 \sum_{\ell=0}^n x^\ell \, dx \\\\ ~~~~~~~~ = \sum_{\ell=0}^n \int_0^1 x^\ell \, dx \\\\ ~~~~~~~~ = \sum_{\ell=0}^n \frac1{\ell+1} \\\\ ~~~~~~~~ = \sum_{\ell=1}^{n+1} \frac1\ell = H_{n+1}[/tex]
where
[tex]H_n = \displaystyle \sum_{\ell=1}^n \frac1\ell = 1 + \frac12 + \frac 13 + \cdots + \frac1n[/tex]
is the [tex]n[/tex]-th harmonic number. The generating function will be useful:
[tex]\displaystyle \sum_{n=1}^\infty H_n x^n = -\frac{\ln(1-x)}{1-x}[/tex]
To evaluate the remaining sum to get [tex]S[/tex], let
[tex]\displaystyle f(x) = \sum_{n=1}^\infty \frac{H_{n+1}}{n(n+1)} x^{n+1}[/tex]
and observe that [tex]S=\lim\limilts_{x\to-1^+} f(x)[/tex], which I'll abbreviate to [tex]f(-1)[/tex]. Differentiating twice, we have
[tex]\displaystyle f'(x) = \sum_{n=1}^\infty \frac{H_{n+1}}n x^n[/tex]
[tex]\displaystyle f''(x) = \sum_{n=1}^\infty H_{n+1} x^n[/tex]
[tex]\displaystyle \implies f''(x) = -\frac{\ln(1-x)}{x^2(1-x)} - \frac1x[/tex]
By the fundamental theorem of calculus, noting that [tex]f(0)=f'(0)=0[/tex], we have
[tex]\displaystyle \int_{-1}^0 f'(x) \, dx = f(0) - f(-1) \implies f(-1) = -\int_{-1}^0 f'(x) \, dx[/tex]
[tex]\displaystyle \int_x^0 f''(x) \, dx = f'(0) - f'(x) \implies f'(x) = -\int_x^0 f''(t) \, dt[/tex]
[tex]\displaystyle \implies S = f(-1) = \int_{-1}^0 \int_x^0 \left(\frac{\ln(1-t)}{t^2(1-t)} + \frac1t\right) \, dt \, dx[/tex]
Change the order of the integration, and substitute [tex]t=-u[/tex].
[tex]S = \displaystyle \int_{-1}^0 \int_{-1}^t \left(\frac{\ln(1-t)}{t^2(1-t)} + \frac1t\right) \, dx \, dt \\\\ ~~~ = - \int_{-1}^0 \left(\frac{(1+t) \ln(1-t)}{t^2(1-t)} + \frac1t + 1\right) \, dt \\\\ ~~~ = -1 - \int_{-1}^0 \left(\left(\frac2{1-t} + \frac2t + \frac1{t^2}\right) \ln(1-t) + \frac1t\right) \, dt \\\\ ~~~ = -1 - \int_0^1 \left(\left(\frac2{1+u} - \frac2u + \frac1{u^2}\right) \ln(1+u) - \frac1u\right) \, du[/tex]
For the remaining integrals, substitute and use power series.
[tex]\displaystyle \int_0^1 \frac{\ln(1+u)}{1+u} \, du = \int_0^1 \ln(1+u) d(\ln(1+u)) = \frac{\ln^2(2)}2[/tex]
[tex]\displaystyle \int_0^1 \frac{\ln(1+u)}u \, du = - \int_0^1 \frac1u \sum_{k=1}^\infty \frac{(-u)^k}k \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~ = - \sum_{k=1}^\infty \frac{(-1)^k}k \int_0^1 u^{k-1} \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~ = - \sum_{k=1}^\infty \frac{(-1)^k}{k^2} = \frac{\pi^2}{12}[/tex]
[tex]\displaystyle \int_0^1 \frac{\ln(1+u) - u}{u^2} \, du = - \int_0^1 \frac1{u^2} \left(\sum_{k=1}^\infty \frac{(-u)^k}k + u\right) \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = -\int_0^1 \frac1{u^2} \sum_{k=2}^\infty \frac{(-u)^k}k \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = - \sum_{k=2}^\infty \frac{(-1)^k}k \int_0^1 u^{k-2} \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = -\sum_{k=2}^\infty \frac{(-1)^k}{k(k-1)} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \sum_{k=1}^\infty \frac{(-1)^k}{k(k+1)} = 1 - 2\ln(2)[/tex]
Tying everything together, we end up with
[tex]S = -1 - \left(2 \cdot \dfrac{\ln^2(2)}2 - 2 \cdot \dfrac{\pi^2}{12} + (1-2\ln(2))\right) \\\\ ~~~ = \boxed{\frac{\pi^2}6 - 2 + 2\ln(2) - \ln^2(2)}[/tex]