This is left as an exercise in Dummit & Foote, so I'd like to verify my understanding.
Let $\varphi:G\to H$ be a homomorphism of groups. Then $\ker\varphi \vartriangleleft G$ and $G/\ker \varphi\cong \mathrm{Im}(\varphi)$.
For any group homomorphism, it is clear to see that the kernel is a subgroup. Let $N=\ker\varphi$, then the left and right cosets of $N$ in $G$ coincide, so $N\vartriangleleft G$. (I have already proven this statement and have no issues with the proof so I am not repeating it here.)
It remains to show that $G/N\cong\mathrm{Im}(\varphi)$. Define $f: G/N\to\mathrm{Im}(\varphi)$ by $f(aN)=\varphi(a)$. Then for $a,b\in G$ we have
\begin{align}
f(aNbN) &= f(aNNb)\\ &= f(aNb)\\ &= f(abN)\\ &= \varphi(ab)\\ &= \varphi(a)\varphi(b)\\ &= f(aN)f(bN)
\end{align}
and $$
f(a^{-1}N) = \varphi(a^{-1}) = \varphi(a)^{-1} = f(aN)^{-1},
$$
so that $f$ is a homomorphism. Now, if $x=\varphi(a)\in\mathrm{Im}(A)$, then $f(aN) = \varphi(a)=x$, so that $f$ is surjective. Finally, if $aN\in\ker f$, that is, $f(aN) = 1$, then $\varphi(a)=1$. But this means that $a\in N$, so that $aN=N$ and $\ker f=N$, hence $f$ is injective. It follows that $f$ is bijective, and therefore is an isomorphism.
As @Shaun pointed out, we still need to show that $f$ is well-defined. Let $a,b\in G$ and assume $aN=bN$. Then $ab^{-1}\in N=\ker f$, that is, $f(ab^{-1}N)=e$, so that $\varphi(ab^{-1})=e$ and $\varphi(a) = \varphi(b)$. This means $f(aN) = f(bN)$, and so $f$ is well-defined.
Best Answer
Your proof is fine, except that you need to prove that $f$ is well-defined. The proof is like proving $f$ is injective, only in reverse.
Assume $aN=bN$. Then $ab^{-1}\in N=\ker f$. That is, $f(ab^{-1}N)=e$, i.e., $\varphi (ab^{-1})=e$, so $\varphi(a)=\varphi (b)$ since $\varphi$ is an homomorphism; thus $f(aN)=f(bN)$. Hence $f$ is well-defined.