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3 fixed typos

If you mean a (real) analytical solution with $y(0)=0$, then the answer is 'no'. If you write this as the problem of looking for integral curves of $\omega = (2x-x^2y)\ dx + y\ dy$ in the $xy$-plane, you'll immediately see that the origin is an isolated singular point of elliptic type (i.e., the eigenvalues of the linearization at $(x,y)=(0,0)$ are pure imaginary). In particular, there is no nontrivial integral curve of $\omega$ that passes through the origin.

All of the integral curves of $\omega$ near the origin spiral around it and converge inward as they go counterclockwise around the origin. To see this, just consider the vector field $$Z = (2x - x^2y)\frac{\partial\ }{\partial y} - y\frac{\partial\ }{\partial x}$$ which is tangent to the integral curves of $\omega$, and consider the convex function $H = 2x^2 + y^2$. The derivative of $H$ with respect to $Z$ is $-2x^2y^2$, which is non-positive and vanishes only on the axes, which are not tangent to $Z$. Since $H$ is strictly decreasing along each integral curve of $Z$, the integral curves of $Z$ spiral in to the origin.

Added note about complex solutions: I neglected to mention that, over the complex numbers, there are, of course, two solutions satisfying $y(0)=0$. I doubt that there is a closed form expression for them in elementary terms, but one easily obtains that they have convergent power series expansions of the form $$y_{\pm}(x) = {} \pm i\sqrt{2}\ x f_0(x^4) + x^3 f_1(x^4),$$ where $$f_0(t) = 1 - \frac{t}{3\cdot2^6} + \frac{13\ t^2}{5\cdot3^2\cdot2^{13}} - \cdots$$ and $$f_1(t) = \frac{1}{2^4} frac{1}{2^2} + \frac{t}{3\cdot2^9} - \frac{11\ t^2}{5\cdot3^3\cdot2^{15}} + \cdots.$$

2 Added information about complex solutions

If you mean an a (real) analytical solution with $y(0)=0$, then the answer is 'no'. If you write this as the problem of looking for integral curves of $\omega = (2x-x^2y)\ dx + y\ dy$ in the $xy$-plane, you'll immediately see that the origin is an isolated singular point of elliptic type (i.e., the eigenvalues of the linearization at $(x,y)=(0,0)$ are pure imaginary). In particular, there is no nontrivial integral curve of $\omega$ that passes through the origin.

All of the integral curves of $\omega$ near the origin spiral around it and converge inward as they go counterclockwise around the origin. To see this, just consider the vector field $$Z = (2x - x^2y)\frac{\partial\ }{\partial y} - y\frac{\partial\ }{\partial x}$$ which is tangent to the integral curves of $\omega$, and consider the convex function $H = 2x^2 + y^2$. The derivative of $H$ with respect to $Z$ is $-2x^2y^2$, which is non-positive and vanishes only on the axes, which are not tangent to $Z$. Since $H$ is strictly decreasing along each integral curve of $Z$, the integral curves of $Z$ spiral in to the origin.

Added note about complex solutions: I neglected to mention that, over the complex numbers, there are, of course, two solutions satisfying $y(0)=0$. I doubt that there is a closed form expression for them in elementary terms, but one easily obtains that they have convergent power series expansions of the form $$y_{\pm}(x) = {} \pm i\sqrt{2}\ x f_0(x^4) + x^3 f_1(x^4),$$ where $$f_0(t) = 1 - \frac{t}{3\cdot2^6} + \frac{13\ t^2}{5\cdot3^2\cdot2^{13}} - \cdots$$ and $$f_1(t) = \frac{1}{2^4} + \frac{t}{3\cdot2^9} - \frac{11\ t^2}{5\cdot3^3\cdot2^{15}} + \cdots.$$

1

If you mean an analytical solution with $y(0)=0$, then the answer is 'no'. If you write this as the problem of looking for integral curves of $\omega = (2x-x^2y)\ dx + y\ dy$ in the $xy$-plane, you'll immediately see that the origin is an isolated singular point of elliptic type (i.e., the eigenvalues of the linearization at $(x,y)=(0,0)$ are pure imaginary). In particular, there is no nontrivial integral curve of $\omega$ that passes through the origin.

All of the integral curves of $\omega$ near the origin spiral around it and converge inward as they go counterclockwise around the origin. To see this, just consider the vector field $$Z = (2x - x^2y)\frac{\partial\ }{\partial y} - y\frac{\partial\ }{\partial x}$$ which is tangent to the integral curves of $\omega$, and consider the convex function $H = 2x^2 + y^2$. The derivative of $H$ with respect to $Z$ is $-2x^2y^2$, which is non-positive and vanishes only on the axes, which are not tangent to $Z$. Since $H$ is strictly decreasing along each integral curve of $Z$, the integral curves of $Z$ spiral in to the origin.
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