#### Answer

$\dfrac{\partial h}{\partial t_2}=0$

#### Work Step by Step

We have $\dfrac{\partial h}{\partial x}=\dfrac{1}{y}; \dfrac{\partial h}{\partial y}=\dfrac{-x}{y^2}; \dfrac{\partial x}{\partial t_2}=t_1; \dfrac{\partial y}{\partial t_2}=t_1^2$
We apply the chain rule to obtain:
$\dfrac{\partial h}{\partial t_2}=\dfrac{\partial h}{\partial x} \dfrac{\partial x}{\partial t_2}+\dfrac{\partial h}{\partial y} \dfrac{\partial y}{\partial t_2} \\=t_1(\dfrac{1}{y})-t_1^2 (\dfrac{x}{y^2})$
We are given that $x=t_1t_2; y=t_1^2t_2$
Thus, $\dfrac{\partial h}{\partial t_2}=t_1(\dfrac{1}{t_1^2t_2})-t_1^2 (\dfrac{t_1t_2}{(t_1^2t_2)^2})$
Therefore, we get $\dfrac{\partial h}{\partial t_2}=0$