Application of Implicit Function Theorem to a function $\psi:U\subset\Bbb{R}^{2}\rightarrow \Bbb{R}^{4}$

Let $$U$$ be an open subset of $$\Bbb{R}^{2}$$ and

\begin{align*}\psi:U&\rightarrow\Bbb{R}^{4}\\ x\mapsto &(\psi_1(x),\psi_2(x),\psi_3(x),\psi_4(x)) \end{align*} a $$\mathcal C^1$$ function.

Assume that there exists $$(\overline{x_1},\overline{x_2})\in U$$ such that $$\text{d}\psi_{(\overline{x_1},\overline{x_2})}$$ is injective.

First, I needed to prove the existence of $$i,j\in\{1,2,3,4\},$$ $$i and an open set $$V\subset U$$ containing $$(\overline{x_1},\overline{x_2})$$, such that, the function $$h:V\rightarrow\Bbb{R}^{2}$$ defined by

$$h(x_1,x_2)=(\psi_i(x_1,x_2),\psi_j(x_1,x_2))$$

is a diffeomorphism of class $$\mathcal C^1$$ into an open set $$W\subset \Bbb{R}^{2}$$.

I have proved the above statement using the inverse function theorem. Note that, if $$\mathrm{d}\psi_{(\overline{x_1},\overline{x_2})}$$ is injective, we can choose two rows $$i,j$$ of this matrix that are linearly independent, so we can apply the inverse function theorem to $$h$$.

But the second question is where I'm stucked:

Consider $$i=1,j=2$$. I need to show that $$\forall$$ $$(x_1,x_2)\in V,$$ $$\exists!$$ $$(y_1,y_2)\in W$$ and a function $$f=(f_1,f_2):W\rightarrow \Bbb{R}^{2}$$ such that $$\psi(x_1,x_2)=(y_1,y_2,f(y_1,y_2)).$$

My problem here is how to prove that $$\dfrac{\partial \psi}{\partial x_3,x_4}$$ is non-singular.

By the same idea of the first question, the linearly independents rows of $$\mathrm{d}\psi_{(\overline{x_1},\overline{x_2})}$$ are $$1$$ and $$2$$.

What can I do?

There is no need for $$\dfrac{\partial \psi}{\partial x_3,x_4}$$ to be non-singular. There are no restrictions on $$\psi$$'s behavior with respect to $$x_3, x_4$$ at all. Using your earlier notation $$(y_1, y_2) = h(x_1, x_2)$$ and $$f(y_1, y_2) = (\psi_3\circ h^{-1}(y_1, y_2),\psi_4\circ h^{-1}(y_1, y_2))$$