Let $M \subset \mathbb{C}^n $ be an algebraic $n-1$-dimensional manifold given as vanishing hypersurface of a polynomial $F \in \mathbb{C}[x_1,..., x_n]$ of degree $d \ge 2$. The smoothness of $M$ can be characterized in pure algebraic terms by requiring that the gradient / normal vector $n_M(p):= (\frac{dF}{dx_1}(p), ..., \frac{dF}{dx_n}(p))$ is not zero for every $p \in M$. Note that since we assumed that $F$ has degree $d \ge 2$, $M$ is not a linear subspace and therefore $n_M(p)$ is not constant. We call $M \to \mathbb{C}^n, \ p \mapsto n_M(p) $ the Gauss map.
Question: Why is it true that the restriction of the gradient $n_M$ to every algebraic curve $C \subset M$, ie $1$-dimensional algebraic submanifold, is not constant? I found a projective form of this claim in J. Harris' First Course in Algebraic Geometry at the end of page 188. There is clamed that the Gauss map
$$ \mathcal{G}_M: M \to \mathbb{P}^{n*}, \ \ p \mapsto [\frac{dF}{dx_1}(p): ... : \frac{dF}{dx_n}(p)]$$
is finite. Obviousy that's equivalent to the claim avove that no curve is contracted to a single point.
What I tried so far: the claim is that to every algebraic curve $C \subset M$ the composition $ C \hookrightarrow M \xrightarrow{n_M} \mathbb{C}^n $ cannot be constant.
Let $p \in M$ and $C$ a curve embedded in $M$ through $p$. Since $C$ is a $1$-algebraic manifold We can $C$ parametrize locally via a chart map $c: (- \epsilon, \epsilon) \to C, t \mapsto c(t):= (c_1(t),..., c_n(t)), c(0)=p$, for appropriate polynomials $c_k(t) \in \mathbb{C}[t]$. Since $ C \subset M$, the composition $F \circ c: t \mapsto F(c(t))$ is zero and therefore it's derivative $(F \circ c)'(t)= F'(c(t)) \cdot c'(t)$ is zero too. Note that $F'(p)= n_M(p)$ and $c'(t)$ is a non-zero tangent vector of $M$ at $c(t)$.
We obtain the non suprising fact that the normal vector $F'(p)=n_M(p)$ of $M$ is perpendicular to tangent vector $c'(t)$ along the curve $c(t)$. Assume that $n_M(p)$ is constant along $c(t)$.
If we can show that every curve $C$ in $M$ through $p$ is not contained it the affine space $p+ T_p M \cong p +\mathbb{C}^{n-1}$, we win but I not see any plausible reason for this to be true.
This is certainly false. Consider the cylinder $x^2+y^2=1$ in $\Bbb C^3$. (Any developable ruled surface will do.)