Question: $$ \frac{x}{{(x^2+y^2)}^{\frac{2}{3}}} + \frac{y}{{(x^2+y^2)}^{\frac{3}{2}}} \cdot \frac{dy}{dx} = 0 $$
First, we need to find $$ M_y(x,y) $$
Where, $$\begin{align*}
M_y(x,y) &= \frac{\partial}{\partial y}M(x,y) \\
&= \frac{\partial}{\partial y}\frac{x}{{(x^2+y^2)}^{\frac{3}{2}}} \\
&= x \cdot \frac{\partial}{\partial y}(x^2+y^2)^{-\frac{3}{2}} \\
&= x \cdot ({-\frac{3}{2}}) \cdot (x^2+y^2)^{-\frac{5}{2}} \cdot 2y \\
&= -\frac{3xy}{(x^2+y^2)^{\frac{5}{2}}}
\end{align*}$$
Then for the $$N_x(x,y)$$
Where, $$\begin{align*}
N_x(x,y) &= \frac{\partial}{\partial x} \\
&= \frac{\partial}{\partial x}\frac{y}{{(x^2+y^2)}^{\frac{3}{2}}} \\
&= y \cdot \frac{\partial}{\partial x}(x^2+y^2)^{-\frac{3}{2}} \\
&= y \cdot ({-\frac{3}{2}}) \cdot (x^2+y^2)^{-\frac{5}{2}} \cdot 2x \\
&= -\frac{3yx}{(x^2+y^2)^{\frac{5}{2}}}
\end{align*}$$
Since $$M_y(x,y) = N_x(x,y)$$ are exact, there exist a Φ, where Φx = M, Φy = N.
Therefore, $$\begin{align*} \phi &= \int \frac{x}{{(x^2+y^2)}^{\frac{2}{3}}} dx \\
&=\frac{-1}{{(x^2+y^2)}^{\frac{1}{2}}} dx + g(y)
\end{align*}$$
$$\phi_y = \frac{y}{{(x^2+y^2)}^{\frac{2}{3}}} + g'(y) $$
Compare it to N(x,y), we know that g'(y) = 0, so g(y) = constant.
Therefore, $$\phi = \frac{-1}{{(x^2+y^2)}^{\frac{1}{2}}} + C $$
General Solution: $$\frac{-1}{{(x^2+y^2)}^{\frac{1}{2}}} = C$$
