This is a nice exercise...Hints
1) $\,\operatorname{Aut}(N)\,$ is abelian
2) Every inner automorphism of $\,G\,$ is, when restricted to $\,N\,$ , is an element of $\,\operatorname{Aut}(N)\,$
3) $\,\forall x,y\in G\,\,,\,[x,y]^{-1}=[y,x]\,$
Try now to do something with this and, if after thinking it over for a while you're still stuck, write back below as a comment.
Added on request: As noted, $\,\operatorname{Aut}(N)\,$ is abelian and if $\,\phi_g\,$ denotes the inner automorphism determined by $\,g\,$ , then $\,\forall\,g\in G\,\,\,,\,\,\text{then}\;\; \left.\phi_g\right|_N\,\in\operatorname{Aut}(N)$ . We show now that any basic commutator $\,[x,y]\in H\,$ centralizes any element $\,n\in N\,$ :
$$[x,y]n[x,y]^{-1}=[x,y]n[y,x]=x^{-1}y^{-1}xyny^{-1}x^{-1}yx=\left(\phi_{x^{-1}}\phi_{y^{-1}}\phi_x\phi_y\right)(n)=$$
$$\stackrel{\text{Aut}(N)\,\,\text{is abelian!}}=\left(\phi_{x^{-1}}\phi_x\phi_{y^{-1}}\phi_y\right)(n)=Id_N\circ Id_N(n)=n$$
and since the above is true for any generator of $\,H=G'=[G,G]\,$ then it is true for the whole group.
Second solution: Perhaps easier: for any subgroup $\,K\leq G\,$ , the map $$f:N_G(K)\to\operatorname{Aut}(K)\,\,,\,\,f(k):=\phi_k=\,\text{conjugation by}\,\,k$$
is a group homomorphism (with $\,\phi_k(x):=kxk^{-1}\,$), whose kernel is precisely $\,C_G(K)\,$ , and from here
$$N_G(K)/C_G(K)\cong T\leq\operatorname{Aut}(K)$$
In our case, we have $\,N\triangleleft G\Longleftrightarrow N_G(N)=G\,$ , so that we get $\,G/C_G(N)\cong T\leq\operatorname{Aut}(N)\,$ .
But $\,\operatorname{Aut}(N)\,$ is abelian, so that
$$G/C_G(N)\,\,\,\text{is abelian}\,\,\Longleftrightarrow G'\leq C_G(N)\;\;\;\;\;\;\square$$
Best Answer
If $g\in G$ and $h\in H$, then $ghg^{-1}h^{-1}=h'$, for some $h'\in H$ (since $H$ contains the commutator subgroup). But then $ghg^{-1}=h'h\in H$. Therefore, $gHg^{-1}\subset H$.