Show that the equation $y^3+ye^{u+v}+2=0$ has a unique solution $y=f(u,v)$ defined for all $(u,v)\in\mathbb R^2$. Conclude that $f$ is continuous everywhere.
Let $G(u,v,y)=y^3+ye^{u+v}+2$ then $G:\mathbb R^{2+1}\rightarrow\mathbb R$ is a $\mathscr C^1$ function and $$D_y(G)=3y^2+e^{u+v}\neq 0$$
OK, then the implicit function theorem ensure there exist a continuous function $f:\mathbb R^2\rightarrow \mathbb R$ such that $y=f(u,v)$. But how could I get $f$ explicitly? Because there was no point given which I can used to get $\{f_n\}_0^\infty$ converges uniformly to $f$.
Another question (didn't mention by @ClaudeLeibovici answer) which bother me a lot is,
How $\mathscr C^1$ of $G$ implies that $f$ is $\mathscr C^1$ in general? I just need the intuition why the fact is true.
I always forget to check the first condition of IFT that, "$\cdots$ in a neighborhood of the point $(\textbf a,\textbf b)$ where $G(\textbf a,\textbf b)=0$ $\cdots$" Like here the implicit function theorem ensure that there exist a continuous function $f:J\subset\mathbb R^2\rightarrow \mathbb R$ centered at $\textbf a\in\mathbf R^2$ such that $y=f(u,v)$. Then how to conclude $f$ is continuous everywhere?
More general question is, how to show such $(\textbf a,\textbf b)$ exist in the context of finding $y=f(u,v)$ defined for all $(u,v)\in\mathbb R^2$ in general (not based on this question)
Let $u+v=k$ and solve the cubic equation $$y^3+e^k\,y+2=0$$ The discriminant $$\Delta=-4 \left(e^{3 k}+27\right)$$ is always negative; so only one real root given by $$y=-\frac 2{\sqrt 3} e^{k/2} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(3 \sqrt{3} e^{-3 k/2}\right)\right)$$ This is obtained using the hyperbolic method.