Real Analysis Question! Consider the equation $\sin(x^2 + y) − 2x= 0$ for $x ∈ \mathbb{R}$ with $y ∈ \mathbb{R}$
as a parameter.
Prove the existence of neighborhoods $V$ and $U$ of $0$ in $\mathbb{R}$ such that for every
$y ∈ V$ there exists a unique solution $x = ψ(y) ∈ U$. Prove that $ψ$ is a $C\infty$
mapping on $V$, and that $ψ'(0) = \frac{1}{2}.$
I know that the solution has to do with the inverse and implicit function theorems but I just can't figure it out! Any help would be much appreciated!

Implicit differentiation of
$$
\sin(x^2+y)-2x=0\tag{1}
$$
yields
$$
y'=2\sec(x^2+y)-2x\tag{2}
$$
$(1)$ implies
$$
|x|\le\frac12\tag{3}
$$
$(2)$ and $(3)$ imply
$$
\begin{align}
|y'|
&\ge2|\sec(x^2+y)|-2|x|\\
&\ge2-1\\
&=1\tag{4}
\end{align}
$$
By the Inverse Function Theorem, $(4)$ says that for all $y$,
$$
|\psi'(y)|\le1\tag{5}
$$
and $\psi\in C^\infty$. Furthermore, $x=0$ is the only $x\in\left[-\frac12,\frac12\right]$ so that $\sin(x^2)=x$. $(2)$ says that $y'=2$ at $(0,0)$. Therefore,
$$
\psi'(0)=\frac12\tag{6}
$$
Since $\psi$ is continuous, there is a neighborhood of $y=0$ so that $\psi'(y)>0$. Thus, $\psi$ is unique in that neighborhood of $y=0$.
