Implicit Differentiation.

$3{ y }^{ 2 }+2{ x }^{ 3 }=9x\\ \frac { d }{ dx } 3{ y }^{ 2 }+\frac { d }{ dx } { 2x }^{ 3 }=\frac { d }{ dx } 9x\\ \frac { d }{ dx } 6y+6{ x }^{ 2 }=9\\ \frac { d }{ dx } 6y=9-6{ x }^{ 2 }\\ \frac { d }{ dx } =\frac { (9-6{ x }^{ 2 }) }{ 6y } \\ \frac { d }{ dx } =\frac { 9 }{ 6y } -\frac { 6{ x }^{ 2 } }{ 6y } \\ \frac { d }{ dx } =\frac { 3 }{ 2y } -\frac { { x }^{ 2 } }{ y }$

Short way to imlicit differentiation, First write the equation in the form $f(x,y)=0$ Then $\frac{dy}{dx}=-\frac{(\frac{df}{dx})}{(\frac{df}{dy})}$

(Here df/dx is partial differentiation meaning differentiation of f taking y as constant & the same goes for df/dy as well.)

Hence from our function, $3y^2+2x^3-9x=0$ $\frac{dy}{dx}=-\frac{6x^2-9}{6y}=\boxed{3/2y-x^2/y}$

$3y^2+2x^3=9x$

By implicit differentiation, we have

$6y\dfrac{dy}{dx}+6x^2=9$

$6y\dfrac{dy}{dx}=9-6x^2$

Dividing both sides by $6y$ , we have

$\dfrac{dy}{dx}=\dfrac{9-6x^2}{6y}=\dfrac{3}{2y}-\dfrac{x^2}{y}$