A way to be certain where to use the definition to evaluate $f_x(x,y)$ and where one may differentiate $f(x,y)$ wrt $x$ mechanically using formulas

Question

Given a function $f(x,y)=xy(\frac{x^2-y^2}{x^2+y^2}),(x,y)\neq (0,0)\in \Bbb R^2$ and $f(0,0)=0.$ Find the value of $f_x(0,0).$

I think something was wrong with this problem. This is because, we have that $$f_x(x,y)=\frac{x^4y+x^2y^3+3x^2y^3-y^5}{(x^2+y^2)^2}.$$ So, it is clearly evident that $f_x(x,y)$ is undefined at $(0,0).$ But later I realized that I was wrong and I had to use the definition of partial derivative of $f$, i.e $$f_x(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}\frac{0-0}{h}=0.$$

But I had one confusion, to which I did not seem to have a good answer. When I worked out, the derivative of $f$ with respect to $x$ as, $f_x(x,y)=\frac{x^4y+x^2y^3+3x^2y^3-y^5}{(x^2+y^2)^2}$, then, it seems I am implicitly assuming that, $f_x(x,y)=\lim_{h\to 0}\frac{f(x+h,y)-f(x,y)}{h}$ and $(x,y)\neq (0,0).$ In general when given an arbitrary function $f(x,y)$ not necessarily this function, that has a single "formula" for calculating the functional values at it's every points in the range of definition, and we are to calculate $\frac{\partial f(x,y)}{\partial x}$ then we can mechanically differentiate $f$ wrt $x$ treating $y$ as a constant using the well known formulas for differentiation like $\frac{d x^n}{dx}=nx^{n-1},\frac{d(y_1+y_2)}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx},$ etc. However, we could not do this with this given $f$ for as I mention, doing that would imply, that I am implicitly assuming that, $f_x(x,y)=\lim_{h\to 0}\frac{f(x+h,y)-f(x,y)}{h}$ and $(x,y)\neq (0,0).$ But the thing is:

Is there a motivation or an intuition so that i can avoid making this mistake in future? I want to know is there any way that while performing a partial differentiation on $f(x,y)$ wrt to an independent variable say, $x$ by which I am supposed to know that differentiating wrt to $x$ using standard formulas mechanically will not serve the purpose and is a wrong step, in there (for instance)? I want a rigorous way to be certain where to use the definition to evaluate $f_x(x,y)$ and where I may directly differentiate $f(x,y)$ wrt $x$ (,treating $y$ as constant) mechanically using the standard formulas.

Answer 1

This is not a question of multivariable calculus. Go back to one variable. If I give you $$f(x)=\begin{cases} g(x), & x\ne a \\ c, & x=a\end{cases},$$ then what is $f’(a)$? Of course, we know that $f’(x)=g’(x)$ whenever $x\ne a$. You have no option but to use the definition of the derivative to calculate $f’(a)$.

Answer 2

Any point $(a,b)$ which is not equal to $(0,0)$ is surrounded by points which are also not equal to $(0,0)$, say a disk centered at $(a,b)$ with small enough radius. At all points $(x,y)$ in this disk, the value of $f(x,y)$ is given by the formula $f(x,y) = xy \, \frac{x^2-y^2}{x^2+y^2}$, and those are the only values that are relevant in the definition of $f_x(a,b)$, so you can differentiate that expression according to the usual rules and evaluate the result at the point $(a,b)$ to obtain $f_x(a,b)$. The special value $f(0,0)=0$ plays no role in this calculation, since the point $(0,0)$ lies outside of the disk.

On the other hand, the partial derivative $f_x(0,0)$ depends both on the value of $f$ at the origin, that is $f(0,0)=0$, and on the values of $f$ at (nearby) points on the $x$-axis, that is $f(x,0)=x \cdot 0 \cdot \frac{x^2-0^2}{x^2+0^2}$ for (small) $x \neq 0$. So you need to fall back to the definition to compute $f_x(0,0)$. The formula which gives $f(x,y)$ for $(x,y) \neq (0,0)$ has no way of “knowing” what the value $f(0,0)$ is, so there's no way that you could find $f_x(0,0)$ by using only that formula. (In particular, computing $f_x(x,y)$ for $(x,y) \neq (0,0)$ from that formula and then taking the limit of the result as $(x,y) \to (0,0)$ will in general not give you the correct value of $f_x(0,0)$.)