Calculation:
Given function:
\( f(x,y) = \begin{cases} \dfrac{xy}{\sqrt{x^2+y^2}}, & (x,y) \ne (0,0) \ 0, & (x,y) = (0,0) \end{cases} \)
Continuity at (0,0)
Put \( x = r\cos\theta,\ y = r\sin\theta \)
\( f(x,y) = \dfrac{r\cos\theta \cdot r\sin\theta}{r} = r\cos\theta \sin\theta \)
\( |f(x,y)| \le r \)
\( \lim_{r \to 0} f(x,y) = 0 = f(0,0) \)
Hence, \( f \) is continuous at \( (0,0) \).
Differentiability at (0,0)
\( f_x(0,0) = \lim_{h \to 0} \frac{f(h,0)-0}{h} = 0 \)
\( f_y(0,0) = \lim_{k \to 0} \frac{f(0,k)-0}{k} = 0 \)
Now check:
\( \lim_{(x,y)\to(0,0)} \frac{f(x,y) - f(0,0)}{\sqrt{x^2+y^2}} \)
\( = \frac{xy}{x^2+y^2} \)
Along \( y = x \):
\( \frac{x^2}{2x^2} = \frac{1}{2} \)
Along \( y = -x \):
\( \frac{-x^2}{2x^2} = -\frac{1}{2} \)
Limits are different, so limit does not exist.
Hence, \( f \) is not differentiable at \( (0,0) \).
\( f_x(0,0) = 0 \) and \( f_y(0,0) = 0 \)
\({\text{Both } f_x \text{ and } f_y \text{ exist at } (0,0)} \)
The correct answer is option 1.