Answer
$$\dfrac{1+x^4y^2+y^2+x^4y^4-2xy}{x^2-2xy-2x^5y^3}$$
Work Step by Step
Re-arrange the equations as: $\tan^{-1} (x^2y)-x-xy^2=0$
Now, $F(x,y)=\tan^{-1} (x^2y)-x-xy^2=0$
and $$F_x=\dfrac{(2xy)}{1+(x^2y)^2} -1-y^2 \\ F_y=\dfrac{x^2}{1+(x^2y)^2} -2xy$$
Since, $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$
Thus,
$$\dfrac{dy}{dx}=-\dfrac{\dfrac{(2xy)}{1+(x^2y)^2} -1-y^2}{\dfrac{x^2}{1+(x^2y)^2} -2xy} \\=\dfrac{(1+y^2)(1+x^4y^2)-2xy}{x^2-2xy(1+x^4y^2)} \\=\dfrac{1+x^4y^2+y^2+x^4y^4-2xy}{x^2-2xy-2x^5y^3}$$
