If $f : X \to Y$ is a diffeomorphism where $y = f(x)$ is a regular value for $f$, can we conclude that $x$ is a regular value for $f^{-1}$?
My attempted proof:
Let $f : X \to Y$ be a diffeomorphism between two smooth manifolds, where $y = f(x)$ is a regular value for $f$.
Since $f$ is a diffeomorphism it has an inverse $f^{-1}$, and since $f$ is bijective we have $f^{-1} \circ f = \text{id}_X$. By the chain rule we get $d(f^{-1} \circ f)_x = df^{-1}_{f(x)} \circ df_x = df^{-1}_{y} \circ df_x$
Suppose now that $x$ was not a regular value for $f^{-1}$, then $df_y$ would be nonsingular, and not of full rank, hence $df^{-1}_{y} \circ df_x$ being the product of two matrices would also not be of full rank, hence $d(f^{-1} \circ f)_x$ is nonsingular implying that $x$ is not a regular value for $\text{id}_X$, a clear contradiction.
Therefore $x$ must be a regular value for $f^{-1}$. $\ \ \square$
Is my proof correct? Is there any easier way to prove this?
Your proof is correct, but there is a much easier way to prove this. A regular value $y$ of a smooth map $f:X\to Y$ is a point such that for every point $x$ in $f^{-1}(\{y\})$, the map $Df_x:TX_x\to TY_y$ is surjective.
Now note that the derivative of any diffeomorphism $f$ is an isomorphism, which is, in particular, surjective. Let $g$ denote the inverse of $f$. Since $y$ is the only point in $g^{-1}(\{x\})$, and $Dg_y$ is an isomorphism, it is a surjection, proving the claim.