consider $f(x,y,z)=x^4+2x\cos y+\sin z$, proof that in a neighborhood of $0$, the equation $f(x,y,z)=0$ sets $z$ as a function of class $C^{\infty}$ of the variables $x,y$. compute $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$.
i set $f(x,y,z)=0$ then
$$\begin{align} f(x,y,z)&=x^4+2x\cos y+\sin z\\ f(x,y,z)&=0\\ x^4+2x\cos y+\sin z&=0\\ \sin z&=-x^4-2x\cos y\\ \frac{\partial}{\partial x}\sin z&=\frac{\partial}{\partial x}\left(-x^4-2x\cos y\right)\\ \cos z\frac{\partial z}{\partial x}&=-4x^3-2\cos y\\ \frac{\partial z}{\partial x}&=\frac{-4x^3-2\cos y}{\cos z}\\ \frac{\partial }{\partial y}\sin z&=\frac{\partial }{\partial y}\left(-x^4-2x\cos y\right)\\ \cos z\frac{\partial z}{\partial y}&=2x\sin y\\ \frac{\partial z}{\partial y}&=\frac{2x\sin y}{\cos z} \end{align}$$
but how i can prove that the function $z$ is function of Class $C^{\infty}$?
For each $(x,y)\in (-1/3,1/3)\times \mathbb {R},$ we have $-x^4 -2x\cos y \in (-1,1).$ Hence for such an $(x,y)$ there is a unique $z\in (-\pi/2,\pi/2)$ such that $\sin z = -x^4 -2x\cos y.$ That $z$ is of course $\arcsin (-x^4 -2x\cos y).$ So there's your function of $(x,y).$ It's a composition of $C^\infty$ functions, and so belongs to $C^\infty ((-1/3,1/3)\times \mathbb {R}).$ (Note: The interval $(-1/3,1/3)$ was just the first thing I saw that would things work out nicely - nothing special about it.)