= Solution
The original PDF has $8y^3-4x$ in the denominator and $dy/dx$ on the left. Rearrange it into the <differential form>
$$
(5x+4y)\,dx+(4x-8y^3)\,dy=0.
$$
The 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 $F=\tfrac52x^2+4xy+h(y)$. Matching $F_y=4x-8y^3$ gives $h(y)=-2y^4$. Therefore an implicit <general solution> is
$$
\boxed{\frac52x^2+4xy-2y^4=C.}
$$
Differentiating verifies $y'=(5x+4y)/(8y^3-4x)$. A solution graph is understood only on intervals where its denominator is nonzero; the printed quotient does not define a <derivative> when $8y^3=4x$.
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