I am trying to figure out why my function is differentiable and therefore continious.
Considering my function: $f(x, y) = x^3 - 3xy^2$
The partiel derivate of $x$ at the position $x,y$ is: $$ \frac{\partial f }{\partial x}(x,y) = 3x^2-3y^2 $$ The partiel derivate of $y$ at the position $x,y$ is: $$ \frac{\partial f }{\partial y}(x,y) = -6xy $$
Therefore I get: $$\nabla f(x,y) = \begin{pmatrix} \frac{\partial f }{\partial x}\\ \frac{\partial f }{\partial y} \end{pmatrix} = \begin{pmatrix} 3x^2-3y^2 \\ -6xy \end{pmatrix} $$
The limit value investigation at a certain point from the left and from the right only makes sense for a function with only one variable.
How can I proof if my function is differentiable?
Let $(a,b)$ be any point in $\mathbb{R^2}$. Starting from where you left off, we need to show $f$ is differentiable at $(a,b)$. We calculate: $f(a+h,b+k) - f(a,b)-((3a^2-3b^2)h+(-6ab)k)= (a+h)^3-3(a+h)(b+k)^2-a^3+3ab^2-3a^2h+3b^2h+6abk=3ah^2-3ak^2-6bhk+h^3-3hk^2$. Thus put $A = \left(3a^2-3b^2, -6ab\right)\implies \dfrac{\left|f(a+h,b+k) - f(a,b) - A\cdot \begin{pmatrix} h\\k \end{pmatrix}\right|}{|(h,k)|}=\dfrac{|3a(h^2-k^2)+ h(h^2-6bk-3k^2)|}{\sqrt{h^2+k^2}}\le \dfrac{|3a||(h^2+k^2)}{\sqrt{h^2+k^2}}+ \dfrac{|h|}{\sqrt{h^2+k^2}}\cdot|h^2-6bk-3k^2|\le |3a|\sqrt{h^2+k^2} + |h^2-6bk-3k^2|\to 0$ when $(h,k) \to (0,0) \implies f'(a,b) = A$ as claimed.