Partial derivative at origin Given $u(x,y)=(x^3-3y^2x)/(x^2+y^2)$, $x,y\in \mathbb R$. Then, partial derivative of $u$ with respect to $x$ at $(0,0)$ i.e. $u_x(0,0)=?$
I calculated as follows: $u_x(0,0)=\lim\limits_{h\to 0}\dfrac{u(0+h,0)-u(0,0)}{h}=\dfrac{h^3/h^2}{h}=1$, and by another way I calculated as follows: Differentiating $u$ partially with respect to $x$, we have $u_x(x,y)=\dfrac{(x^2+y^2)(3x^2-3y^2)-(x^3-3y^2x)2x}{(x^2+y^2)^2}$, and then $u_x(0,0)=0$. Why are the two answers different?
 A: The first approach is correct and it gave you the right answer.
Using the second approach, you did not get $0$. What you got was $\frac00$, which is meaningless.
A: To expand a little on José's answer:
The expression for $u$ that you gave in the question only makes sense for $(x,y) \neq (0,0)$, and it must be supplemented with a definition of $u(0,0)$ for the question to be meaningful to begin with. (The natural value to use is the one making $u$ continuous at the origin, namely $u(0,0)=0$, as you have implicitly done in your first approach.)
In your second approach, you never use the value of $u(0,0)$. What you get by differenting the expression for $u(x,y)$ for $(x,y) \neq (0,0)$ is (of course) only the value of $u_x(x,y)$ for $(x,y) \neq (0,0)$, which really tells you nothing at all about $u_x(0,0)$. (You can't count on getting the right value of $u_x(0,0)$ by taking the limit of the expression for $u_x(x,y)$ as $(x,y) \to (0,0)$, since you don't know in advance whether $u_x$ is continuous at the origin or not.)
