Why is the function differentiable in the point $(0,0)$? I am trying to figure out why my function is differentiable and therefore continuous in the point $(0,0)$ which is also a critical point and a saddle point.
Considering my function: $f(x, y) = x^3 - 3xy^2$
The partial derivative of $x$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial x}(x,y) = 3x^2-3y^2 $$
The partial derivative of $y$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial y}(x,y) = -6xy $$
Therefore I get: $$\nabla f(x,y) = \begin{pmatrix} \frac{\partial f }{\partial x}\\ \frac{\partial f }{\partial y} \end{pmatrix} = \begin{pmatrix} 3x^2-3y^2 \\ -6xy  \end{pmatrix} $$
How can I show that my function is point-wise differentiable in $(0,0)$?
 A: Since the partial derivatives exist and are continuous at every point, then the function is differentiable at every point.
A: Besides from the solution proposed by @SonGohan, you can prove it by the definition.
The candidate to be the derivative is given by $L(x,y) = (3x^{2} - 3y^{2},-6xy)$.
Now you can prove that
\begin{align*}
\lim_{(x,y)\to(0,0)}\frac{|f(x,y) - f(0,0) - L(x,y)((x,y) - (0,0))^{T}|}{\|(x,y)\|} = 0
\end{align*}
Indeed, this is the case.
To begin with, notice the numerator can be written as
\begin{align*}
x^{3} - 3xy^{2} - (3x^{3} - 3xy^{2} - 6xy^{2}) = -2x^{3} - 6xy^{2}
\end{align*}
Thence we conclude that the proposed limit equals zero indeed.
Such claim results from the convenient application of the squeeze theorem.
To comprehend this, let us notice that
\begin{align*}
0 < x^{2} \leq x^{2} +y^{2} & \Rightarrow 0 < |x| \leq \sqrt{x^{2} + y^{2}}\\\\
& \Rightarrow 0 < \frac{|x|}{\sqrt{x^{2}+y^{2}}} \leq 1\\\\
& \Rightarrow 0 < \frac{|x|^{3}}{\sqrt{x^{2}+y^{2}}} \leq |x|^{2}\\\\
& \Rightarrow 0 < \left|\frac{x^{3}}{\sqrt{x^{2}+y^{2}}}\right| \leq |x^{2}|
\end{align*}
Similarly, we have that
\begin{align*}
0 < y^{2} \leq x^{2} + y^{2} & \Rightarrow 0 < |y| \leq \sqrt{x^{2}+y^{2}}\\\\
& \Rightarrow 0 < \frac{|y|}{\sqrt{x^{2}+y^{2}}} \leq 1\\\\
& \Rightarrow 0 < \frac{|xy^{2}|}{\sqrt{x^{2}+y^{2}}} \leq |xy|\\\\
& \Rightarrow 0 < \left|\frac{xy^{2}}{\sqrt{x^{2}+y^{2}}}\right| \leq |xy|
\end{align*}
Finally, since we also have
\begin{align*}
0 < \left|\frac{2x^{3} + 6xy^{2}}{\sqrt{x^{2}+y^{2}}}\right| \leq \left|\frac{2x^{3}}{\sqrt{x^{2}+y^{2}}}\right| + \left|\frac{6xy^{2}}{\sqrt{x^{2}+y^{2}}}\right| \leq |2x^{2}| + |6xy|
\end{align*}
we are done.
Hopefully this helps!
A: This is just a polynomial in $x$ and $y$, so it's a sum of products of infinitely differentiable functions starting with the projections $p(x,y) = x$ and $q(x,y) = y$. That makes it infinitely differentiable.
