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how would I calculate the Hessian matrix at the origin for the function $f$ defined as: $f \left( x,y\right) :=\dfrac {\left( x^{3}y-xy^{3}\right) } {x^{2}+y^{2} }$ whenever $(x,y) \neq(0,0)$ and $f(x,y)=(0,0)$ if $(x,y)=(0,0)$

I normally know how to calculate the hessian matrix of a function, but this has confused me because the function is made up of two functions. Any help would be appreciated :)

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    You can proceed as usual, except for the point $(0,0)$, where you have to calculate $f_{xx}$, $f_{xy}$ and $f_{yy}$ using the definition. – 2017-02-27

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For example:

$$f'_x(0,0):=\lim_{x\to0}\frac{f(x,0)-f(0,0)}x=\lim_{x\to0}\frac{0-0}x=0$$

and simmilarly $\;f'_y(0,0)=0\;$ .

Also

$$f''_{xx}(0,0)=\lim_{x\to0}\frac{f'(x,0)-f'(0,0)}x=0$$

because

$$f'_x(x,y)=\frac{(x^2+y^2)(3x^2y-y^3)-(x^3y-xy^3)2x}{(x^2+y^2)^2}\implies f'_x(x,0)=\frac{x^2\cdot0-0\cdot2x}{x^4}=0$$

and etc.

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    Thank you! I think I understand it a lot more. Does that mean the matrix is the zero matrix? – 2017-02-27
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    @Shaunzander Looks like, but you have first to evaluate all the partial derivatives at the origin – 2017-02-27
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One of the clasical method for obtaining the limit of function with two variables in $(0,0)$ is by this fact that you can assume $x=r \, \cos(\theta)$ and $y=r \, \sin(\theta)$, then calculate the limit with new values for $x$ and $y$ when $r \to 0$, if after calculation limit, the final values of limit depend on to $\theta$, we conclude that, the function dose not have limit in $(0,0)$ but if the final values of limit be independent of $\theta$, we say the function has limit in $(0,0)$ with value that we obtained. so we have

$$\lim_{(x,y) \to (0,0)}\,\dfrac {\left( x^{3}y-xy^{3}\right) } {x^{2}+y^{2} } = \lim_{r \to 0}\,\dfrac {\left( {(r \, \cos(\theta))}^{3}\,{r \, \sin(\theta)}-{r \, \cos(\theta)}\,{(r \, \sin(\theta))}^{3}\right) } {{(r \, \cos(\theta))}^{2}+{(r \, \cos(\theta))}^{2} } $$

$$ =\lim_{r \to 0} \frac{r^4({\cos(\theta)^3\, \sin(\theta)}-{\sin(\theta)^3\, \cos(\theta)})}{r^2} $$

$$ =\lim_{r \to 0} r^2\,{(\cos(\theta)^3\, \sin(\theta)}-{\sin(\theta)^3\, \cos(\theta)})=0 $$ The last equation holds because the multi of $0$ in the bounded function is zero.