Diffeomorphism from level set onto $S^2$ I'm given a map $\phi : R^4 \rightarrow R^2$ defined by $\phi(x,y,s,t) = (x^2 + y, x^2+y^2+s^2+t^2+y)$.

It's easy to show that the level set $C = \phi^{-1}(0,1)$ is a smooth submanifold of $R^4$ with dimension $2$ by showing that $(0,1)$ is a regular value. 

I'm trying to show that this level set is diffeomorphic to $S^2$, the unit sphere in $R^3$. The only thing I can think of is to construct a smooth embedding from $S^2$ into $R^4$ whose image is $C$, since smooth embeddings are diffeomorphisms onto their images. 

It's easy enough to get that $C$ is diffeomorphic to $\{(x,y,z) \mid x^4 + y^2 + z^2 = 1\}$, but from there I'm stuck. The map $(x, y, z) \rightarrow (\text{sign}(x) x^2, y, z)$ is a differentiable homeomorphism from $S^2$ onto $C$, but it's not even $C^2$ let alone smooth.
Once you're working with the surface $C=\{ (x,y,z) : x^4+y^2+z^2=1 \}$, how about the map 
$$(x,y,z) \mapsto \frac{(x,y,z)}{ \sqrt{x^2+y^2+z^2}} ?$$

$C$ is a ball-type surface centered at the origin, so why not just project to the sphere in this simple way?
The level set is traced out by the intersection of x^2+y=0 and x^2 + y^2 +s^2 +t^2 + y =1. Combine the equations and get y^2 + s^2 + t^2 = 1.