Answer
$ \frac{dz}{dt}= 3 y^2 x^2 \frac{dx}{dt}+ 2x^3 y\frac{dy}{dt}$
$ \frac{dz}{dt}=-12 <0 \Longrightarrow z$ is decreasing
Work Step by Step
$ z=x^3y^2$
$ \frac{dz}{dt}= \frac{dz}{dx} \frac{dx}{dt}+ \frac{dx}{dy}\frac{dy}{dt}$ ........... eq(1)
Putting $ z=x^3y^2$ in RHS of equation (1)
$ \frac{dz}{dt}=\frac{d(x^3y^2)} {dx} \frac{dx}{dt}+ \frac{d(x^3y^2)}{dy}\frac{dy}{dt}$
$ \frac{dz}{dt}= y^2 \frac{d(x^3)} {dx} \frac{dx}{dt}+ x^3\frac{d(y^2)} {dy}\frac{dy}{dt}$
$ \frac{dz}{dt}= y^2 (3x^2) \frac{dx}{dt}+ x^3 (2y) \frac{dy}{dt}$
$ \frac{dz}{dt}= 3 y^2 x^2 \frac{dx}{dt}+ 2x^3 y\frac{dy}{dt}$ .......... eq (2)
Put $ x=1$,$y=2$, $ \frac{dx}{dt}=-2$ units/s $ \frac{dy}{dt}=3$ units/s in equation (2)
$ \frac{dz}{dt}= 3 \times 2^2 \times (1^2) \times (-2) + 2 \times (1^3) \times 2 \times 3 = -12$.
Since $ \frac{dz}{dt}=-12 <0 \Longrightarrow z$ is decreasing
