Answer
$1$ and $0$
Work Step by Step
We know that $f_x(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0+h,y_0)-f(x_0,y_0)}{h}$
$f_x(0,0)=\lim\limits_{h \to 0} \dfrac{f(0+h,0)-f(0,0)}{h}=\lim\limits_{h \to 0} \dfrac{\frac{\sin (h^3+0^4)}{h^2+0^2}}{h}=1$
Also, $f_y(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0,y_0+h)-f(x_0,y_0)}{h}$
Now,
$f_y(0,0)=\lim\limits_{h \to 0} \dfrac{f(0,0+h)-f(0,0)}{h}=\lim\limits_{h \to 0} \dfrac{\frac{\sin (0^3+h^4)}{0^2+h^2}}{h}=\lim\limits_{h \to 0} \dfrac{\sin h^4}{h^4 } \lim\limits_{h \to 0} (h) =1 \times 0=0$
