Suppose $\varphi:\mathbb{R}^2 \to \mathbb{R}^3$ is defined as:
$$\varphi(u,v)=(u-\frac{u^3}{3}+uv^2,v-\frac{v^3}{3}+vu^2,u^2-v^2)$$
Can I conclude that $\varphi$ is an homeomorphism with its image?
I tried defining $x=u-\frac{u^3}{3}+uv^2$, $y=v-\frac{v^3}{3}+vu^2$ and $z=u^2-v^2$ finding $u=\frac{3x}{3-z}$ and $v=\frac{3y}{1+z}$ but I do not know if this is enough to conclude.
No. Your function is not injective, since$$\varphi\left(\sqrt3,0\right)=\varphi\left(-\sqrt3,0\right)=(0,0,3).$$