Diffeomorphism $f:U\to V$ then $Df:\mathbb{R^n}\to\mathbb{R^m}$ will be linear isomorphism hence $n=m$

$f\circ f^{-1}=Id, f^{-1}\circ f=Id$ implies $Df\circ Df^{-1}=Id, Df^{-1}\circ Df=Id$. It implies $Df$ is an isomorphism.

Hint: What do you get when you apply the chain rule to $$f \circ f^{-1} = \text{Id}?$$

Given $a \in U$, $f \circ f^{-1} = I_{\mathbb R^m}$, so as $f^{-1}$ is differentiable at $f(a)$, $Df(f^{-1}(f(a))) \circ Df^{-1} (f(a)) = I_{\mathbb R^m}$, i.e. $Df(a) \circ Df^{-1} (b) = I_{\mathbb R^m}$, where $b = f(a)$. Similarly, $Df^{-1}(b) \circ Df(a) = I_{\mathbb R^n}$, so $Df(a)$ is invertible and $[Df(a)]^{-1} = Df^{-1}(b)$; it is an isomorphism since it is linear and continuous with a continuous inverse by the assumptions in the question.