Partial Derivatives of Discontinuous Fcn?

[absci2010]

f(x,y)= xy²/(x²+y²) if (x,y)\neq(0,0)
=0 if (x,y)=(0,0)
Show that the partial derivatives of x and y exist at (0,0).

This may be a really stupid question, but would the partial derivatives of x and y at (0,0) just be 0? I tried taking that partial derivatives of xy²/(x²+y²) and got:
df/dx=[(x²+y²)(y²)-xy²(2x)]/(x²+y²)²
and
df/dy=[(x²+y²)(2xy)-xy²(2y)]/(x²+y²)²
which i don't believe cancel out.
Please help?

 

[HallsofIvy]

You are asked for the partial derivatives at (0, 0) so use the definition directly:

\displaytype\frac{\partial f}{\partial x}(0, 0)= \lim_{h\to 0}\frac{f(h, 0)- f(0, 0)}{h}
\displaytype = \lim_{h\to 0}\frac{\frac{h(0)}{h^2+ 0}}{h}= \lim_{h\to 0} 0= 0

Why do you say this function is discontinuous?

 

[absci2010]

Thanks so much!
I realized after I posted the question that it's not discontinuous. Oops. >.<