 Solution

The original PDF has $8y^3-4x$ in the denominator and $dy/dx$ on the left. Rearrange it into the differential formThe cross partial derivatives of its coefficients are both four, so it is an exact differential form. Integrating its first coefficient in $x$ gives a potential $F=\frac{5}{2}{x}^2+4xy+h(y)$. Matching $F_y=4x-8y^3$ gives $h(y)=-2y^4$. Therefore an implicit general solution isDifferentiating verifies $y^{\prime }=(5x+4y)/(8y^3-4x)$. A solution graph is understood only on intervals where its denominator is nonzero; the printed quotient does not define a derivative when $8y^3=4x$.