According to Jacobi's Triple Product Identity $$\prod_{n \geqslant 1}\left(1+a x^{2 n-1}\right)\left(1+a^{-1} x^{2 n-1}\right)\left(1-x^{2 n}\right)=\sum_{-\infty}^{\infty} a^{n} x^{n^{2}}$$
Now how from this statement the follwing statement comes $$\left(a-a^{-1}\right) \prod_{n \geqslant 1}\left(1-a^{2} x^{n}\right)\left(1-a^{-2} x^{n}\right)\left(1-x^{n}\right)=\sum_{-\infty}^{\infty}(-1)^{n} a^{2 n+1} x^{\frac {n^{2}+n}{2}}$$ And also then by differentiating with respect to $a $ and putting $a=1$ and dividing by 2 $$\prod_{n\geq1}\left(1-x^{n}\right)^{3}=\frac{1}{2} \sum_{-\infty}^{\infty}(-1)^{n}(2 n+1) x^{\frac {n^{2}+n}{2}}$$
I tried by substituting $a $ and $n $ but was not able to reach this form.
Look at the summation side to find the required substitution. $$\sum_{n=-\infty}^\infty(-1)^na^{2n+1}x^{(n^2+n)/2} =a\sum_{n=-\infty}^\infty(-a^2x^{1/2})^n(x^{1/2})^{n^2}.$$ So you need to replace $a$ by $-a^2x^{1/2}$ and $x$ by $x^{1/2}$ in the original triple product formula to get this.
You can write the LHS of your second equation in the form $$g(a)=(a-a^{-1})f(a)$$ whose derivative is $$g'(a)=(a-a^{-1})f'(a)+(1+a^{-2})f(a).$$ Then $$g'(1)=2f(1),$$