Let $M,N$ be two complex manifolds, $\emptyset\neq X\subset M\times N$ and let $\pi:M\times N\rightarrow M$ be a projection.
Take $\Gamma\subset X$ - submanifold such that $\dim\Gamma =\dim X$. How do we know that $\pi |_{\Gamma}:\Gamma\rightarrow M$ is differentiable and holomorphic map between manifolds?
Set $$\Gamma_0=\left\{ z\in\Gamma \ : \ rk_z(\pi|_{\Gamma})=rk(\pi|_{\Gamma})\right\},$$
where $rk(\pi|_{\Gamma})=\max\left\{rk_x(\pi|_{\Gamma}), \ x\in\Gamma\right\}$ and the rank at point $x$ is $\dim\text{Im}\ d_x(\pi|_{\Gamma})$. How do we know that $\Gamma_0$ is open in $\Gamma$?
When your write
you presumably assume that it is a complex submanifold. This means that $\Gamma$ is a complex manifold itself and the embedding $j \colon \Gamma \hookrightarrow X$ is holomorphic.
Thus $\pi|_{\Gamma} = \pi \circ j$ is a composition of holomorphic functions and hence holomorphic (and differentiable).
To see that the locus where the rank of $D\pi|_{\Gamma}$ is maximal is open one can apply the following argument:
Let $Z \subset X$ be the set $Z:=\Gamma \setminus \Gamma_0$. It is sufficient to prove that $\Gamma$ is locally closed. For each point $z \in Z$ choose local holomorphic coordinates for both neighbourhoods of $z$ and $\pi|_{\Gamma}(z)$. Then locally the differential $D\pi|_{\Gamma}$ can be represented as a matrix with coefficients in holomorphic functions.
Recall that for a given matrix $A = (a_{i,j})$ and an integer $r$ the condition $\operatorname{rk}(A) < r$ is given by algebraic functions on the coefficients $a_{i,j}$. Therefore, locally $Z$ is defined by a number of algebraic equations on local holomorphic coordinates, hence it is locally a proper complex subvariety --- particularly it is locally closed.
There is also an argument which does not use local coordinates: the differential $D\pi|_{\Gamma}$ gives a holomorphic section of the bundle $Hom(T\Gamma, \pi|_{\Gamma}^*TM)$. Let $T$ be the total space of this bundle. Then we obtain a holomorphic (indeed only continuity is important) map $s \colon \Gamma \to T$. Let $R_p \subset T$ be the subvariety of those maps $A_z \colon T_{z}\Gamma \to \pi|_{\Gamma}^*T_{\pi|_{\Gamma}(z)}M$ for which $rk(A_z) < p$. It is easy to verify that all $R_p$ are closed and hence $Z = s(\Gamma) \bigcap R_{\mathrm{rk}_z\pi|_{\Gamma}}$ is closed.