How do you show that these equations

$\displaystyle x^2 - y\cos(uv)+z^2 =0 $

$\displaystyle x^2 + y^2 - \sin(uv)+2z^2 =2 $

$\displaystyle xy-\sin(u) \cos(v) +z=0$

determine x,y and z as functions of (u,v) when (u,v,x,y,z) is close to $\displaystyle (\frac{\pi}{2}, 0,1,1,0)$