Let $f:\mathbb R^n\longrightarrow \mathbb R$. Why to define the tangent plane at $a\in\mathbb R^n$ we need that $f$ is differentible at $a$ (and not only partially differentiable) ?
Indeed, if it's partially differentiable, I can define $$z=f(a)+\nabla f(a)\cdot (x-a)$$ that exist since partial derivatives exists. And this is the equation of the tangent plan, no ?
Consider $f: \mathbb R^2 \to \mathbb R$ defined by $$f(x,y) = \begin{cases} x \ \ \ {\rm if \ } x=y \\ 0 \ \ \ {\rm if \ } x \neq y\end{cases}$$ Can you visualise this?
$f$ has well-defined partial derivatives at $(0,0)$, namely, $$\frac{\partial f }{\partial x}|_{(0,0)} = \frac{\partial f }{ \partial y}|_{(0,0)} = 0.$$
However, $f$ is not differentiable. For example, consider the line parameterised by $(x,y) = (t,t)$, with $t \in \mathbb R$. Along the direction of this line, the directional derivative is $$ \frac {df}{dt}|_{t = 0} = 1 \neq 0 = \frac{\partial f }{\partial x}|_{(0,0)} \frac{dx}{dt}|_{t = 0} +\frac{\partial f }{ \partial y}|_{(0,0)}\frac{dy}{dt}|_{t = 0}, $$ Thus the partial derivatives fail to reproduce the slope of $f$ along the direction of the line $(x,y) = (t,t)$.
From the partial derivatives, you might initially guess that $z = 0$ is some sort of tangent plane to the graph of $z = f(x,y)$ at $(0,0)$. But the plane $z = 0$ is not a good candidate for a tangent plane because it does not lie tangent to $z = f(x,y)$ in the direction of the line $(x,y)= (t,t)$.