Unsolvable Differential Equation?

Is there an analytic solution to the following equation?

$$ \frac{dx}{dy} = \frac{x^2 - y^2}{x^2 + y^2} $$

I believe the answer is 'no'. It isn't separable or exact I've had trouble finding any documentation that can help me explore this problem further. I have solved the problem numerically to explore the solution space, and it seems to make sense as the answer, but I also need to show some work in attempting to solve it analytically.

Is there a substitution that can be done to solve this? Or is this, in fact, an unsolvable problem (analytically)?

The following procedure gives more geometrical insight, but in the end leads to cumbersome calculations as well. Put $$x:=r(\phi)\cos\phi,\quad y:=r(\phi)\sin\phi$$ with the polar angle $\phi$ as new independent variable and unknown function $\phi\mapsto r(\phi)$. Then $$x'=r'\cos\phi-r\sin\phi,\qquad y'=r'\sin\phi+r\cos\phi\ .$$ In this way we obtain $${r'\cos\phi-r\sin\phi\over r'\sin\phi+r\cos\phi}={dx\over dy}={x^2-y^2\over x^2+y^2}=\cos(2\phi)\ ,$$ or $${r'\over r}={\sin\phi+\cos\phi\cos(2\phi)\over\cos\phi-\sin\phi\cos(2\phi)}=:f(\phi)\ .$$ Determine a primitive $F$ of $f$ in a maximal interval containing $\phi=0$, and you obtain a global description of the solution curves.

Given homogenius ODE is $$\dfrac{\mathrm dy}{\mathrm dx} = -\dfrac{y^2+x^2}{y^2-x^2}.$$ Can be applied substitution $$y= (u-1)x,\tag1$$ $$x\dfrac{\mathrm du}{\mathrm dx} = -u+1-\dfrac{(u-1)^2+1}{(u-1)^2-1},$$ $$x\dfrac{\mathrm du}{\mathrm dx}=-u-\dfrac{2}{(u-1)^2-1},$$

$$\dfrac{\mathrm dx}{x}=\dfrac{2u-u^2}{u^3-2u^2+2}{\mathrm du}.\tag2$$ Since the denominator of RHS(2) has a root $$r=\dfrac13(2-\sqrt[3]{19+\sqrt{297}}-\sqrt[3]{19-\sqrt{297}})\approx-0.83928\,67552\,14161,\tag3$$ \begin{align} &27(r^3-2r^2+2) = (3r-2)^3 - 12(3r-2) + 38\\ &= -\left(\sqrt[3]{19+\sqrt{297}\,}\,+\sqrt[3]{19-\sqrt{297}\,}\,\right)^3 +12\left(\sqrt[3]{19+\sqrt{297}}+\sqrt[3]{19-\sqrt{297}}\right)+38\\ &= \left(-3\sqrt[3]{19+\sqrt{297}\,}\,\sqrt[3]{19-\sqrt{297}\,}+12\right)\left(\sqrt[3]{19+\sqrt{297}\,}\,+\sqrt[3]{19-\sqrt{297}\,}\,\right)\\ &= \left(-3\sqrt[3]{64}+12\right)\left(\sqrt[3]{19+\sqrt{297}\,}\,+\sqrt[3]{19-\sqrt{297}\,}\,\right) = 0, \end{align} then $$u^3-2u^2+2 = (u-r)(u^2-(2-r)u+r^2-2r),$$ $$R(u)=\dfrac{2u-u^2}{u^3-2u^2+2}=\dfrac A{u-r}+\dfrac{B(2u-2+r)}{u^2-(2-r)u+r^2-2r} +\dfrac{C}{u^2-(2-r)u+r^2-2r},$$ where $$A = \lim_{u\to r}(u-r)R(u) = \dfrac{2r-r^2}{3r^2-4r} = -\dfrac{2-r}{4-3r},$$ $$A+2B = \lim_{u\to \infty}uR(u) = -1,\quad B=-\dfrac{1-r}{4-3r},$$ $$\dfrac {B-A}r+\dfrac C{r^2-2r} = R(0)=0,\quad C=\dfrac{2-r}{4-3r},$$ $$u^2-(2-r)u+r^2-2r = \left(u-\dfrac{2-r}2\right)^2+\dfrac{(2-r)(-3r-2)}4.$$

Then from $(2)$ should $$(4-3r)\ln \mathrm{const}|x| = -(2-r)\ln|u-r|-(1-r)\ln|u^2-(2-r)u+r^2-2r|\\ +2\sqrt{\dfrac{2-r}{-3r-2}}\arctan\dfrac{2u-2+r}{\sqrt{(2-r)(-3r-2)}},$$ $$\color{green}{\mathbf{\small C_1|x|^{4-3r}|u-r||u^3-2u^2-2|^{1-r} = \exp\left(2\sqrt{\dfrac{2-r}{-3r-2}}\arctan\dfrac{2u-2+r}{\sqrt{(2-r)(-3r-2)}}\right)}},\tag4$$ where $C_1$ is the arbitrary constant.

Formulas $(1),(3),(4)$ present the common solution of the given ODE.