By using a suitable substitution, bring the equation…
By using a suitable substitution, bring the equation $\dfrac{3y^2y'}{16x}=2xy^3$ to a form with separated variables and integrate it, given that $y(x)$ remains bounded as $x\to\infty$.
Explanation of Method
Cancel the common factor $y^2$ to reduce to a simple separable linear-in-$y$ equation, solve it, then use the boundedness requirement (rather than a point condition) to pin down the constant.
Step 1: Simplify and separate
Multiplying by $\dfrac{16x}{y^2}$ (for $y\neq0$) gives $3y'=32x^2y$, so
\[\frac{dy}{y}=\frac{32x^2}{3}\,dx,\qquad y\neq0.\]
Step 2: Integrate
\[\ln|y|=\frac{32}{9}x^3+C \quad\Longrightarrow\quad \boxed{\,y=Ce^{\frac{32}{9}x^3}.}\]
Step 3: Apply the boundedness condition
For any $C\neq0$, $e^{32x^3/9}\to\infty$ as $x\to\infty$, so $|y|\to\infty$. Boundedness therefore forces $C=0$.
Result
Final answer:
\[\boxed{\,y=Ce^{\frac{32}{9}x^3}\ \text{(general)},\qquad y\equiv0\ \text{is the only solution bounded as } x\to\infty.}\]