Let $f(x) = x^2e^{x^2}$, and assume that $(e^x)' = e^x$ for all $x$ in $R$.
a) Prove that $f^{-1}$ exists and is differentiable on $(0, ∞)$.
Proof: Suppose that $f(x) = x^2e^{x^2}$, then finding the derivative $f'(x) = 2xe^{x^2}(x^2 + 1)$. Then $f'(x) > 0$ when $x>0$ so f is strictly increasing by the Inverse Function Theorem, hence 1-1 on the $(0, ∞)$ for all $x$ in $(0, ∞)$. By the inverse Function Theorem, if $f'(a)$ exists and is nonzero, then $f^{-1}$ is differentiable. Then we conclude $f^{-1}$ is differentiable on $(0, ∞)$.
Can someone please verify this is enough for this proof. If not, can someone please give me feedback. Thanks.
You are right that if $f'(x)>0$ for all $x$ and $f'$ is continuous implies that $f$ is injective (1-1) on $(0,\infty)$, but for $f^{-1}$ to exist $f$ needs to be also surjective (consider for example $x\mapsto (1-e^{-x})$). Since $f(x)\to \infty$ as $x\to \infty$, and $f(0)=0$, $f$ is surjective.
Otherwise your proof is OK.
Just remember that the inverse function theorem guarantees only a local inverse function, not global. In this case we can show that a global inverse exists and then use the inverse function theorem to prove that this inverse is differentiable.