Let $P$ be a polynomial in $\mathbb{C}[z_1,z_2,...,z_n].$ Consider the derivative of $P,$ $D_{\mathbb{C}}P$, as a holomorphic map from $\mathbb{C}^n$ to $\mathbb{C}^n.$ I have the following question:
Does the condition $P$ being irreducible imply that the set $Z(D_{\mathbb{C}}P)\cap Z(P)$ is atmost finite, where $Z$ represents the zero set?
The answer is no.
Consider the set $X$ given by the zeroes of $P$. This is an algebraic variety in $\mathbb C^n$. The set you write down is the singular locus of $X$.
The singular locus of a variety can be a divisor in your variety, even if your variety is irreducible. In particular, if $\dim X >1$, then the singular locus can be positive-dimensional.
Let me try an explicit example.
We will need a polynomial in at least three variables. Let $P =x^2 z + y^3+y^2$. Note that $P$ is irreducible. To see this, consider $P$ as a linear polynomial in $z$ with coefficients in $\mathbb C[x,y]$.
We want to check that the singular locus of this surface is a curve.
The derivatives of $P$ are $2xz$, $3y^2 + 2y$ and $x^2$. (Explicitly: $dP/dx = 2xz$, $dP/dy = 3y^2 +2y$ and $dP/dz = x^2$.)
Note that if $x = y =0$ all derivatives vanish. Also $P$ vanishes if $x = y =0$. Therefore, the singular locus of $X = Z(P)$ contains the line $z =0$. In your terminology, the set $\{(0,0,z) \ | \ z\in \mathbb C\}$ is contained in $Z(P) \cap Z (DP)$. Thus the singular locus of $X = Z(P)$ is one-dimensional (and infinite).