= Solution
Here the dependent variable is $x$ and the independent variable is $y$. Multiplying the equation by $2x$ and setting $q=x^2$ gives a linear equation:
$$
\frac{dq}{dy}-\frac4yq=2y.
$$
An <integrating factor> is $y^{-4}$, so
$$
\frac d{dy}(q y^{-4})=2y^{-3},\qquad
q y^{-4}=-y^{-2}+C.
$$
Thus $x^2=Cy^4-y^2$. The initial point $(x,y)=(1,1)$ fixes $C=2$, and the positive branch through that point is
$$
\boxed{x(y)=y\sqrt{2y^2-1},\qquad y>\frac1{\sqrt2}.}
$$
Equivalently, inverting this branch gives
$$
\boxed{y(x)=\frac12\sqrt{1+\sqrt{1+8x^2}},\qquad x>0.}
$$
The source equation contains $xy$ in the denominator. The endpoint with $x=0$ is therefore excluded even though the inverse expression has a finite limit there. A negative-$x$ branch cannot be joined through that endpoint to produce a solution $x(y)$ of the original equation. The interval shown is the maximal real positive branch containing the initial point.
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