[tex]v=\dfrac h{3x}\implies\dfrac{\mathrm dv}{\mathrm dt}=\dfrac{3x\frac{\mathrm dh}{\mathrm dt}-3h\frac{\mathrm dx}{\mathrm dt}}{3x^2}[/tex]
By the chain rule,
[tex]\dfrac{\mathrm dh}{\mathrm dt}=\dfrac{\mathrm dh}{\mathrm dx}\dfrac{\mathrm dx}{\mathrm dt}[/tex]
We have
[tex]h(x)=4e^{2x-6}-x^2+5\implies\dfrac{\mathrm dh}{\mathrm dx}=8e^{2x-6}-2x[/tex]
and we're given that [tex]x[/tex] changes at a constant rate of [tex]\frac{\mathrm dx}{\mathrm dt}=0.2[/tex] thousand people per minute, which means
[tex]\dfrac{\mathrm dv}{\mathrm dt}=\dfrac{3x\left(8e^{2x-6}-2x\right)\left(0.2\frac{\text{thousand people}}{\rm min}\right)-3h\left(0.2\frac{\text{thousand people}}{\rm min}\right)}{3x^2}[/tex]
At the moment [tex]x=4[/tex] thousand people are in the park, we have [tex]h(4)=4e^2-11[/tex], so we find
[tex]\dfrac{\mathrm dv}{\mathrm dt}=\dfrac{12\left(8e^2-8\right)-3(4e^2-11)}{240}\dfrac{\text{thousand people}}{\rm min}=\dfrac{7(4e^2-3)}{80}\dfrac{\text{thousand people}}{\rm min}[/tex]
or approximately 2.324 thousand people per minute.
