The total derivatives are: $$dz=(\frac{\partial z}{\partial x})_ydx+(\frac{\partial z}{\partial y})_xdy$$ $$dx=\frac{dx}{dt}dt$$ $$dy=\frac{dy}{dt}dt$$ Substituting into the first equation gives: $$dz=(\frac{\partial z}{\partial x})_y\frac{dx}{dt}dt+(\frac{\partial z}{\partial y})_x\frac{dy}{dt}dt$$
is it acceptable to simply divide by $dt$ now? Or is there a more correct way of obtaining $\frac{dz}{dt}$
Just use the chain rule $$ \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}. $$
Proof. Suppose $t$ changes by $\Delta t$. Then $x,y$ changes by $\Delta x$ and $\Delta y$ and $z$ changes by $\Delta z$ and $$ \Delta z=\frac{\partial z}{\partial x}\Delta x+\frac{\partial z}{\partial y}\Delta y+\epsilon_1\Delta x+\epsilon_2\Delta y, $$ where $\epsilon_{1,2}\to0$ as $\Delta x,\Delta y\to0$. Then $$ \frac{\Delta z}{\Delta t}=\frac{\partial z}{\partial x}\frac{\Delta x}{\Delta t}+\frac{\partial z}{\partial y}\frac{\Delta y}{\Delta t}+\epsilon_1\frac{\Delta x}{\Delta t}+\epsilon_2\frac{\Delta y}{\Delta t}. $$ If $x(t), y(t)$ differentiable at $t$, then $$ \lim_{\Delta t\to 0}\frac{\Delta x}{\Delta t}=\frac{dx}{dt}, \lim_{\Delta t\to 0}\frac{\Delta y}{\Delta t}=\frac{dy}{dt}, $$ hence $$ \lim_{\Delta t\to0}\frac{\Delta z}{\Delta t}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}. $$ $z(x(t),y(t))$ is differentiable at $t$.