Question concerning the surface $\phi(u, v)=(u, v^3, u-v)$ Let $\phi:R^2 \rightarrow R^3$ ,$C^{\infty}$ with $\phi(u,v)=(u,v^3,u-v)$. And $\gamma(t)=(3t,t^6,3t-t^2)$ smooth curve .
Prove:
$\textbf{a)}$ $M=\phi(R^2)$ is a smooth surface.
$\textbf{b)}$  that $\gamma(R) \subset M$. Which is the $C^{\infty}$ parametrized curve $\phi^{-1} \circ \gamma$
$\textbf{c)}$ express the velocity $\dot{\gamma(0)}$ as a linear compination of the vectors of the basis of the tangent plane $T_0M$ at $\gamma(0)=(0,0,0)$.
Solution:
$\textbf{a)}$ $\phi$ is smooth so only think that remains is to prove that it is an acceptable parametrization for its image. It is 1-1 as its components is 1-1. Onto into its image .its Jacobian matrix has rank 2 . Since its collumns are $(1,0,1),(0,3v^2,-1)$ which with Gauss ellimination even if $v=0$ still there are 2 independent collumns. Only think left is to prove that its inverse IS continuous.$\textbf{ How do i find its inverse?}$
$\textbf{b)}$ Let $t \in R$ then for every $(3t,t^6,3t-t^2)$ i can find $(u(t),v(t))$ such that  $\phi(u(t),v(t)=(3t,t^6,3t-t^2)$ $u(t)=3t$,$ v(t)=t^2$ so every point of the curve is on $M$. I think my justification is awkward how would i write that clearer and more rigorously mathematically??. To find the $\phi^{-1}\circ \gamma$ $\textbf{i need to find the inverse from a).}$
$\textbf{c)}$ is pure calculations it is easy.$d\phi/du(0,0)=(1,0,1)$ $d\phi/dv(0,0)=(0,0,-1)$. $\dot{\gamma(0)}=(3,0,3)=3(d\phi/du)+0(d\phi/dv)$ Are my calculations right?
 A: For a) notice that $\phi(\mathbb{R}^2)=\lbrace (x,y,z)\in\mathbb{R}^3: z=x-\sqrt[3]{y}\rbrace$ hence $\phi(\mathbb{R}^2)$ is the graph of the function $f:\mathbb{R}^2\to\mathbb{R}$ given by $f(x,y)=x-\sqrt[3]{y}$. One can easily show that graphs of smooth functions are always manifolds. 
For b), let $C(t)=(3t,t^2)$. It is clear that $C$ gives a smooth curve in $\mathbb{R}^2$. Notice that $\gamma(t)=\phi(C(t))$. Since $\phi$ is a diffeomorphism and $C$ is a smooth curve, $\gamma$ must also be a smooth curve. That $\gamma\subset M$ comes from $\gamma(t)=\phi(C(t))$.
Claim: Suppose $f:\mathbb{R}^n\to\mathbb{R}$ is a smooth function and let $\Gamma(f)=\lbrace (x_1,\dots,x_n,f(x_1,\dots,x_n)):(x_1,\dots,x_n)\in\mathbb{R}^n\rbrace$, called the Graph of $f$. Then 
 $\Gamma(f)\cong_{diff}\mathbb{R}^n$
Proof: Let $x=(x_1,\dots,x_n)$ for short and let $\phi:\mathbb{R}^n\to \Gamma(f)$ be given by $\phi(x)=(x,f(x))$. Then $\phi$ is smooth as  map between euclidean spaces. Now let $\Phi:\Gamma(f)\to\mathbb{R}^n$ be given by projection onto the first $n$ coordinates restricted to $\Gamma(f)$. Again $\Phi$ is smooth as a map between euclidean spaces. It is easy to show that $\phi\circ\Phi=\text{id}_{\Gamma(f)}$ and $\Phi\circ\phi=\text{id}_{\mathbb{R}^n}$. So $\phi$ is a diffeomorphism and this completes the argument.
EDIT: So in your case, the inverse $\Phi:M\to\mathbb{R}^2$ given by the formula $\Phi(x,y,z)=(x,\sqrt[3]{y})$ will work.
