For Instance: Given the function: $\varphi(x) = x\,\varphi(x)^2+2\,x^2\,e^{\varphi(x)}$.
It has to be shown that it exist a function $\varphi: U \to\mathbb{R}$ that solves it.
In this case $f(x,y) = x\,y^2+2\,x^2\,e^{y}-y\quad$ hence $D_yf(x_0,y_0) = -1$ where $(x_0,y_0) = (0,0)$.
So it is solvable.
Now I ask myself how a function $\varphi(x)$ that depends on more variables $y_1,y_2$ would look like.
In general if there is a function $f(x,y_1,y_2) = \left(\begin{array}{c}x+y_1+{y_2}^2\\x^2-y_2 \end{array}\right)$. How to state a function $\varphi(x)$ ?
For me, this means rearranging for y. But if there are 2 of them? Something like $\varphi(x)=\left(\begin{array}{c}y_1 \\y_2 \end{array}\right)$ ?
Dini's theorem in higher dimension states that, given a function $\underline f:\Omega\subseteq\mathbb R^m\times\mathbb R^n\to\mathbb R^n$, where $\Omega$ is an open set and $\underline f\in\mathcal C^1(\Omega)$, if $(\underline x_0,\underline {y_0})\in\Omega $ such that $\underline f(\underline x_0,\underline {y_0})=0$ and $J\underline f|_{\underline y}(\underline x_0,\underline {y_0})$ is invertible$\implies$
$\exists U(\underline x_0)$ and $V(\underline f(\underline{y_0}))$, with $\overline{U(\underline x_0)\times V(\underline f(\underline{y_0}))}\subseteq\Omega$ such that
$$(1)\text{ The set }Z_{\underline f}=\{\underline x\in\mathbb \Omega\:\underline f(\underline x)=\underline 0\}\text{ resticted to }U(\underline x_0)\times V(\underline f(\underline{y_0}))$$$$\text{ coincides with the graph of the implicit function }\underline \psi: U(\underline x_0)\to V(\underline f(\underline{y_0}))$$ $$(2)\text{ }\underline\psi\in\mathcal C^1(U(\underline x_0))\text{ and }J\underline\psi|_{\underline y}(\underline x)=-(J\underline f|_{\underline y})^{-1}(J\underline f|_{\underline x})(\underline x,\underline\psi(\underline x))\forall \underline x\in U(\underline x_0)$$
In order to see how it works, consider the function $G:\mathbb R^5\to\mathbb R^2$ such that $$G(u,v,x,y,z)=\begin{pmatrix}ux^2-yz^2+v\cos y-1\\xe^y+vz-\cos u-2 \end{pmatrix}$$ and try to verify that $G(u,v,x,y,z)=(0,0)$ implicitly defines a function $(u,v)=\psi (x,y,z)$ in a neighbourhood of the point $(0,1,2,0,1)$.
Hint: if $JG=\begin{pmatrix}x^2&\cos y&2ux&-z^2-v\sin y&-2yz\\\sin u&z&e^y&xe^y&v \end{pmatrix}$ how can you see if $G$ satisfies the Hp of Dini's theorem?