Given the nonlinear system of equation: $$2x+y^3+u^3-v^2=1 \\ x^2+3y-u^2-v^3=0$$ for $z=(x,y,z,v)\in \mathbb{R}^4$ and with a solution $z_0=(1,0,0,1)\in \mathbb{R}^4$, one can conclude that by the implicit function theorem there exisits $\varepsilon , \delta >0$ and a $C^1$-function $g: U_\varepsilon (0,1)\to U_\delta (1,0)$ such that for $(x,y,u,v)\in U_\varepsilon (0,1)\times U_\delta (1,0)$, we have:
$(x,y,u,v)$ solves the system of equation iff $(x,y)=g(u,v)$
Is it possible to locally solve the system of equation at $z_0$ with $(x,u)$ as a function of $(y,v)$?
I hope I get it right, but you're question just seems to consider a re-ordering of your variables. Say $$f(x,y,u,v) = \begin{pmatrix}2x+y^3+u^3-v^2-1\\ x^2+3y - u^3-v^3\end{pmatrix}.$$ If you want to express the variables $(x,u)$ as a function of $(y,v)$ you need to show that the matrix $$ \begin{pmatrix}\partial_y f_1 & \partial_v f_1 \\ \partial_y f_2 & \partial_v f_2\end{pmatrix}=\begin{pmatrix}3y^2 & -2v \\ 3 & -3u^2\end{pmatrix} $$ is invertible at the point $(1,0,0,1)$ (which is on the $0$-level set of $f$), hence you look at $$ \begin{pmatrix}0 & -2 \\ 3 & 0\end{pmatrix}, $$ wich is clearly invertible.