Linear approximation for an implicit function Let $$F(r,\mu,t)=r- \frac{2\mu\left ( \dfrac{e^t(t-1)-1}{t^2} \right )}{(1-2\mu)+2\mu \left ( \dfrac{e^t-1}{t} \right )}=0,$$
I would like to find a linear approximation$$
t(r,\mu) \approx t(0,0) + r\frac{\partial t(r,\mu)}{\partial r}+\mu \frac{\partial t(r,\mu)}{\partial \mu},$$
where$$
\frac{\partial t(r,\mu)}{\partial r}=-\left.\frac{\partial F(r,\mu,t)}{\partial r}\middle/\frac{\partial F(r,\mu,t)}{\partial t}\right., \quad \frac{\partial t(r,\mu)}{\partial \mu}=-\left.\frac{\partial F(r,\mu,t)}{\partial \mu}\middle/\frac{\partial F(r,\mu,t)}{\partial t}\right..$$
I want to find $t(0,0)$, which I do not know how to find it because if I have $t(r,\mu)$ I do not need any more linear approximation.
Any idea or hint for a better approximation of $t(r,\mu)$ would help me a lot.
 A: $\def\e{\mathrm{e}}$Define$$
G(r, μ, t) = rt ((1 - 2μ)t + 2μ (\e^t - 1)) - 2μ ((t - 1)\e^t - 1), %\tag{1}
$$
then $F(r, μ, t) = 0 \Leftrightarrow G(r, μ, t) = 0$. Note that\begin{align*}
\frac{\partial G}{\partial r}(r, μ, t) &= t ((1 - 2μ)t + 2μ (\e^t - 1)),\\
\frac{\partial G}{\partial μ}(r, μ, t) &= -2rt + 2rt (\e^t - 1) - 2((t - 1) \e^t - 1),\\
\frac{\partial G}{\partial t}(r, μ, t) &= 2r (1 - 2μ)t + 2rμ ((t + 1)\e^t - 1) - 2μ t\e^t.
\end{align*}
Suppose $t = f(r, μ)$ satisfies $G(r, μ, t) \equiv 0$, then from (1) there is\begin{align*}
0 &= \frac{\partial}{\partial r}(G(r, μ, f(r, μ))) = \frac{\partial G}{\partial r}(r, μ, f(r, μ)) + \frac{\partial G}{\partial t}(r, μ, f(r, μ)) · \frac{\partial f}{\partial r}(r, μ),\\
0 &= \frac{\partial}{\partial μ}(G(r, μ, f(r, μ))) = \frac{\partial G}{\partial μ}(r, μ, f(r, μ)) + \frac{\partial G}{\partial t}(r, μ, f(r, μ)) · \frac{\partial f}{\partial μ}(r, μ).
\end{align*}
For specific $t_0 = f(r_0, μ_0)$, if $\dfrac{\partial G}{\partial t}(r_0, μ_0, t_0) \neq 0$, then by the implicit function theorem, there is a uniquely-defined $f$ in a neighborhood of $(r_0, μ_0)$ such that $f(r_0, μ_0) = t_0$ and$$
\frac{\partial f}{\partial r}(r_0, μ_0) = -\frac{\dfrac{\partial G}{\partial r}(r_0, μ_0, t_0)}{\dfrac{\partial G}{\partial t}(r_0, μ_0, t_0)}, \quad \frac{\partial f}{\partial μ}(r_0, μ_0) = -\frac{\dfrac{\partial G}{\partial μ}(r_0, μ_0, t_0)}{\dfrac{\partial G}{\partial t}(r_0, μ_0, t_0)}.
$$
Now, note that when $(r_0, μ_0) = \left( \dfrac{1}{2}, \dfrac{1}{2} \right)$,$$
G\left(\frac{1}{2}, \frac{1}{2}, t\right) = 0 \Longleftrightarrow \frac{1}{2}t (\e^t - 1) - ((t - 1) \e^t - 1) = 0 \Longleftrightarrow (t - 2)(\e^t + 1) = 0,
$$
thus $t_0 = 2$ is the unique solution. Because$$
\frac{\partial G}{\partial t}\left( \frac{1}{2}, \frac{1}{2}, 2 \right) = -\frac{1}{2}(\e^2 + 1) \neq 0,
$$
then\begin{align*}
\frac{\partial f}{\partial r}\left( \frac{1}{2}, \frac{1}{2} \right) &= -\frac{\dfrac{\partial G}{\partial r}\left( \dfrac{1}{2}, \dfrac{1}{2}, 2 \right)}{\dfrac{\partial G}{\partial t}\left( \dfrac{1}{2}, \dfrac{1}{2}, 2 \right)} = 4 · \frac{\e^2 - 1}{\e^2 + 1},\\
\frac{\partial f}{\partial μ}\left( \frac{1}{2}, \frac{1}{2} \right) &= -\frac{\dfrac{\partial G}{\partial μ}\left( \dfrac{1}{2}, \dfrac{1}{2}, 2 \right)}{\dfrac{\partial G}{\partial t}\left( \dfrac{1}{2}, \dfrac{1}{2}, 2 \right)} = -\frac{4}{\e^2 + 1},
\end{align*}
and in a neighborhood of $(r_0, μ_0) = \left( \dfrac{1}{2}, \dfrac{1}{2} \right)$, there is\begin{align*}
f(r, μ) &\approx 2 + \frac{\partial f}{\partial r}\left( \frac{1}{2}, \frac{1}{2} \right) \left( r - \frac{1}{2} \right) + \frac{\partial f}{\partial μ}\left( \frac{1}{2}, \frac{1}{2} \right) \left( μ - \frac{1}{2} \right)\\
&= 2 + 4 \left( \frac{\e^2 - 1}{\e^2 + 1}  \left( r - \frac{1}{2} \right) - \frac{1}{\e^2 + 1} \left( μ - \frac{1}{2} \right) \right)\\
&= \frac{2}{\e^2 + 1} (2(\e^2 - 1) r - 2μ + 3).
\end{align*}
