critical points of a map Let $\mathcal{C}$ be the $\mathbb{R}^2$ submanifold defined by:
$y^2 = x(x − 1)(x − a)$ where $a>1$.
Let $f: \mathcal{C} \rightarrow \mathbb{R}^2: (x,y) \rightarrow x$
Find the critical point of $f$.
We have $\partial_x f=1$ and  $\partial_y f=0$, which means that f has no critical points because its gradient is non zero.
On the other hand, I know that $f$ has 3 critical points in $(0,0), (1,0), (a,0)$. How to explain this contradiction?
Thank you for your help!
 A: Let $g\colon\mathbb{R}^2\to\mathbb{R}$ defined by $g(x,y)=x$.
What you have shown is that $g$ has no critical point, however $f=g_{\vert C}$ may have critical points!
Keep in mind that:

A regular map $h\colon M\rightarrow N$ restricted to a submanifold $S$ of $M$ may no longer be regular.

Example. The height function on $\mathbb{R}^3$ has no critical point, however its restriction to the unit sphere $S^2$ has two critical points, the north pole and the south pole, at those points the tangent plane of $S^2$ has constant altitude.

The general phenomena is that at a point $p\in M$, the linear map $T_ph\colon T_pS\to T_{f(p)}N$ may no longer be surjective, since $T_pS$ does not contain all the directions of $T_pM$.

When $N=\mathbb{R}$, we are looking at points $p$, where $T_ph(T_pS)=\{0\}$, since $T_ph$ is a linear form and it is surjective if and only if nonzero.

In your case, you are looking at point $(x,y)\in C$ such that $T_{(x,y)}C$ is contained in a $x$-slice of $\mathbb{R}^2$, this happens at the points $(0,0)$, $(1,0)$ and $(a,0)$.
Since $C$ is defined by the equation $\varphi(x,y)=y^2-x(x-1)(x-a)=0$, the tangent line of $C$ at $(x,y)$ is given by $\nabla_{(x,y)}\varphi^\perp$, so that $T_{(x,y)}C$ is contained in a $x$-slice if and only if $\nabla_{(x,y)}\varphi$ has no $y$-component.
Compute $\nabla_{(x,y)}\varphi=(-3x^2+2(a+1)x-a,2y)$, it has no $y$-component if and only if $y=0$.
Finally, the critical points of $f$ are the $(x,0)\in C$, but $(x,0)\in C$ if and only if $x=0,1,a$. 
Here is a sketch, where $C$ is drawn in red:

