Answer
The equation
$$
\left( \frac{x}{(x^{2}+y^{2})^{\frac{3}{2}}}\right) d x+\left( \frac{y}{(x^{2}+y^{2})^{\frac{3}{2}}}\right)d y=0
$$
is exact, and its solutions are given implicitly by
$$
x^{2}+y^{2} =c
$$
where $ c $ is an arbitrary constant
Work Step by Step
$$
\left( \frac{x}{(x^{2}+y^{2})^{\frac{3}{2}}}\right) d x+\left( \frac{y}{(x^{2}+y^{2})^{\frac{3}{2}}}\right)d y=0 \quad \quad (i)
$$
Comparing this Equation with the differential form:
$$
M(x,y) d x+N(x,y) d y=0
$$
we observe that
$$
\begin{aligned}M(x,y) &= \frac{x}{(x^{2}+y^{2})^{\frac{3}{2}}}
\\ N(x,y) &=\frac{y}{(x^{2}+y^{2})^{\frac{3}{2}}} \end{aligned}
$$
By calculating $M_{y}$ and $N_{x}$ , we find that
$$
M_{y}(x, y)=-3xy(x^{2}+y^{2})^{\frac{-5}{2}}=N_{x}(x, y)
$$
so the given equation is exact. Thus there is a $\psi (x, y)$ such that
$$
\begin{aligned} \psi_{x}(x, y) &=M(x,y) =\frac{x}{(x^{2}+y^{2})^{\frac{3}{2}}} ,
\\ \psi_{y}(x, y) &=N(x,y) =\frac{y}{(x^{2}+y^{2})^{\frac{3}{2}} } \end{aligned} \quad \quad (ii)
$$
Integrating the first of these equations with respect to $x$ , we obtain
$$
\begin{aligned} \psi(x, y) &=\int{\frac{x}{(x^{2}+y^{2})^{\frac{3}{2}}}} dx,
\\ \ & =\frac{1}{2}\int{2x(x^{2}+y^{2})^{\frac{-3}{2}}} dx,
\\ \ & =- (x^{2}+y^{2})^{\frac{-1}{2} } +h(y) \end{aligned} \quad \quad (ii)
$$
where $h(y)$ is an arbitrary function of $y$ only. To try to satisfy the second of Eqs. (ii), we compute $\psi_{y}(x, y) $ from Eq. (iii) and set it equal to $N$, obtaining
$$
\psi_{y}(x, y)=y(x^{2}+y^{2})^{\frac{-3}{2}} +h^{\prime}(y)=\frac{y}{(x^{2}+y^{2})^{\frac{3}{2}} }
$$
Thus $h^{\prime}(y)=0 $ and $h(y)=0 $ The constant of integration can be omitted since any solution of the preceding differential equation is satisfactory.
Now substituting for h(y) in Eq. (iii) gives
$$
\psi(x, y)=- (x^{2}+y^{2})^{\frac{-1}{2} }
$$
Hence solutions of Eq. (i) are given implicitly by
$$
\psi(x, y) =- (x^{2}+y^{2})^{\frac{-1}{2} } =c
$$
rewriting $\psi(x, y)$ as follow
$$
\psi(x, y) = (x^{2}+y^{2}) =c
$$
where $ c $ is an arbitrary constant