Answer
$$
e^{u^{2}-v}-v=1;
$$
(a)
$$
\begin{aligned}
\frac{d u}{d v} &=\frac{1+e^{u^{2}-v}}{2 u e^{u^{2}-v}} . \end{aligned}
$$
(b)
$$
\begin{aligned}
\frac{d v}{d u} &=\frac{2u e^{u^{2}-v}}{e^{u^{2}-v}+1}.
\end{aligned}
$$
Work Step by Step
$$
e^{u^{2}-v}-v=1;
$$
(a) $\frac{d u}{d v}$
Now, we can calculate $\frac{d u}{d v}$ by implicit differentiation,
$$
\begin{aligned} \frac{d}{d v}\left(e^{u^{2}-v}-v\right) &=\frac{d}{d v}(1) \\
\frac{d}{d v}\left(e^{u^{2}-v}\right)-\frac{d}{d v} v &=\frac{d}{d v}(1) \\ e^{u^{2}-v} \frac{d}{d v}\left(u^{2}-v\right)-1 &=0 \\ e^{u^{2}-v}\left(2 u^{2}-v\right)-1 &=0 \\ 2 u \frac{d u}{d v}-1 &=1 \\
\frac{d u}{d v} &=\frac{1}{2 u}\left(\frac{1}{e^{u^{2}-v}}+1\right) \\ \frac{d u}{d v} &=\frac{1+e^{u^{2}-v}}{2 u e^{u^{2}-v}} . \end{aligned}
$$
$$
e^{u^{2}-v}-v=1;
$$
(b) $\frac{d v}{d u}$
$$
\begin{aligned} \frac{d}{d u}\left(e^{u^{2}-v}-v\right) &=\frac{d}{d u}(1) \\
\frac{d}{d u}\left(e^{u^{2}-v}\right)-\frac{d}{d u} v &=\frac{d}{d u}(1) \\
e^{u^{2}-v} \frac{d}{d u}\left(u^{2}-v\right)-\frac{d v}{d u} &=0 \\ e^{u^{2}-v}\left(2 u-\frac{d v}{d u}\right)-\frac{d v}{d u} &=0 \\ e^{u^{2}-v} \frac{d v}{d u}-\frac{d v}{d u} &=-2 u e^{u^{2}-v} \\-\frac{d v}{d u}\left(e^{u^{2}-v}+1\right) &=-2 u e^{u^{2}-v} \\ \frac{d v}{d u} &=\frac{2u e^{u^{2}-v}}{e^{u^{2}-v}+1}.
\end{aligned}
$$
