There are some test for differentiability. Like checking continuous partial derivatives (that implies differentiability). But this method not always works (in many cases works) because is only enought not necessary condition.
Ex: Check that $f(x,y)=x^2y^3$ is differentiable in $(0,0)$.
Solution: If you take partial derivatives you obtain $f_x(x,y)=2xy^3$ which is continuous in $(0,0)$, and $f_y(x,y)=3x^2y^2$ which is continuous in $(0,0)$. Therefore $f$ is differentiable in $(0,0)$.
If you want to prove differentiability in a general context you must use the usual definition, always works but can be difficult in rare cases.
For continuous differentiability you need to check continuity of all partial derivatives of all orders. Is a advantage when you can find a some particular property that all partial derivatives have, for example all are composition of continuous functions (in this case you only need to check if some derivative can have a singular point). You can prove this fact in a particular case only using the chain rule and product derivative properties.
Sometimes partial derivatives can be 0 from some order. Or can have so similar values like trigonometric functions in such case derivatives only differs for a constant or continuous composition.