Question: Suppose $2$ integrating factors of $\frac{dy}{dx} = f(x,y)$ are $\mu_1 = \frac{1}{x^2+y^2}$ and $\mu_2 = xy$. Find $f(x,y)$.
Suppose that $g(x,y)dx + h(x,y)dy = 0$, i.e. $f(x,y) = -\frac{g(x,y)}{h(x,y)}$. Then $$\frac{d}{dy}\frac{g(x,y)}{x^2+y^2} = \frac{d}{dx}\frac{h(x,y)}{x^2+y^2}$$
and
$$\frac{d}{dy}\{xyg(x,y)\} = \frac{d}{dx}\{xyh(x,y)\}.$$
However it seems not so easy to simplify. How to proceed to find $f(x,y)$?
Starting from your calculus which is correct, $$\begin{cases} \frac{\partial}{\partial y}\frac{g}{x^2+y^2} = \frac{\partial}{\partial x}\frac{h}{x^2+y^2}\qquad (1)\\ \frac{\partial}{\partial y}xyg = \frac{\partial}{\partial x}xyh \qquad (2) \end{cases}$$ From $(1)$ : $$\frac{1}{x^2+y^2}\frac{\partial g}{\partial y} -\frac{2y}{(x^2+y^2)^2}g = \frac{1}{x^2+y^2}\frac{\partial h}{\partial x} -\frac{2x}{(x^2+y^2)^2}h $$ $$\frac{\partial h}{\partial x}-\frac{\partial g}{\partial y} =2\frac{xh-yg}{x^2+y^2} \tag 3 $$ From $(2)$ : $$xy\frac{\partial g}{\partial y}+xg = xy\frac{\partial h}{\partial x}+yh $$ $$\frac{\partial h}{\partial x}-\frac{\partial g}{\partial y} =\frac{xg-yh}{xy}\tag 4$$ From $(3)$ and $(4)$ : $$2\frac{xh-yg}{x^2+y^2}=\frac{xg-yh}{xy}$$ $$2\frac{x-y\frac{g}{h}}{x^2+y^2}=\frac{x\frac{g}{h}-y}{xy}$$ Since $\quad f=-\frac{g}{h}$ $$2\frac{x+yf}{x^2+y^2}=\frac{-xf-y}{xy}$$ Solving for $f\:$ leads to : $$f(x,y) = -\frac{y\,(3x^2+y^2)}{x\,(x^2+3y^2)}$$