#### Answer

\begin{align*}
\frac{\partial r}{\partial t} &= \frac{ e^{s / r} }{2r+\frac{st}{r^2}e^{s/r} }\\
\frac{\partial t}{\partial r} &= \frac{2r+\frac{st}{r^2}e^{s/r} }{ e^{s / r} }
\end{align*}

#### Work Step by Step

Given $$r^{2}=t e^{s / r}$$
Consider $$F(r,s,t )=r^{2}-t e^{s / r} $$
Then
\begin{align*}
F_{r}&=2r+\frac{st}{r^2}e^{s/r}\\
F_{t}&=- e^{s / r}
\end{align*}
Then
\begin{align*}
\frac{\partial r}{\partial t}&=-\frac{F_{t}}{F_{r}}\\
&= \frac{ e^{s / r} }{2r+\frac{st}{r^2}e^{s/r} }
\end{align*}
and
\begin{align*}
\frac{\partial t}{\partial r}&=-\frac{F_{r}}{F_{t}}\\
&= \frac{2r+\frac{st}{r^2}e^{s/r} }{ e^{s / r} }
\end{align*}