Answer
$1, 0$
Work Step by Step
Since, $f_x(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0+h,y_0)-f(x_0,y_0)}{h}$ and $f_y(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0,y_0+h)-f(x_0,y_0)}{h}$
$f_x(0,0)=\lim\limits_{h \to 0} \dfrac{f(0+h,0)-f(0,0)}{h}=\lim\limits_{h \to 0} \dfrac{\dfrac{\sin (h^3+0)}{h^2+0}}{h}=1$
and $f_y(0,0)=\lim\limits_{h \to 0} \dfrac{\dfrac{\sin (0+h^4)}{0+h^2}}{h}$
or, $=\lim\limits_{h \to 0} \dfrac{\sin h^4}{h^4 } \lim\limits_{h \to 0} (h) =0$
