$\displaystyle f(x,y)=\frac{x^{4}}{x^{4}+y^{4}}$ if $\displaystyle (x,y)\ne(0,0)$
$\displaystyle f(x,y)=0$ if $\displaystyle (x,y)=(0,0)$
I have no idea how to start it, thank you very much!
You would start by trying to differentiate it! I assume you know that a function of two variables is differentiable at a point if its partial derivatives exist and are continuous in some neighborhood of that point.
It should be clear that there is no problem anywhere except possibly at (0,0). What are the partial derivatives? Are they continuous at (0,0)?
There is no problem when $\displaystyle x^{2}=-y^{2}$?
To be able to differentiate it must be defined in the point and be continuous, it is defined so we have to watch the limits:
When $\displaystyle x=0 \rightarrow f(x,y)=\frac{0}{y^{4}}=0$
When $\displaystyle y=0 \rightarrow f(x,y)=\frac{x^{4}}{x^{4}}=1$
As $\displaystyle 0\ne1$ there is no limit, so the function is no differentiable in (0,0). Right?
x and y are both real numbers. Theirs squares are both non-negative so $\displaystyle x^2$ can't be equal to $\displaystyle -y^2$!
To be able to differentiate it must be defined in the point and be continuous, it is defined so we have to watch the limits:
When $\displaystyle x=0 \rightarrow f(x,y)=\frac{0}{y^{4}}=0$
When $\displaystyle y=0 \rightarrow f(x,y)=\frac{x^{4}}{x^{4}}=1$
As $\displaystyle 0\ne1$ there is no limit, so the function is no differentiable in (0,0). Right?[/QUOTE]
Yes, excellent point!