Assume $k>0$ is a parameter, and the function $f(x)$ is given by $$f(x)=e^{kx}-ke^{-x}-k+1.$$ How can I calculate the $x_0$ such that $f(x_0)=0$?
First of all, I let $y=e^{x}$, then the equation becomes $f(y)=y^k-ky^{-1}-k+1$. It is worthing noting that when $y\rightarrow 0$, $f(y) \rightarrow -\infty$. When $y\rightarrow +\infty$, $f(y) \rightarrow +\infty$. Thus, there exists at least one $y_0$ such that $f(y_0)=0$. However, I do not konw how to find the closed-form of the solution.
This is not an answer.
As said in comments, there is no hope for a closed form and numerical methods should be used. More than likely, Newton method would be the simplest; however, as usual, the problem is the initial guess $x_0$.
Considering the case where $k >1$ for which the solution is $>0$, I tried to generate some estimate building the $[1,n]$ Padé approximant of the original function. This gives (this is just an example for $n=5$) $$x_0=\frac{12 (k-1) \left(k^8+16 k^7+41 k^6+36 k^5+52 k^4+36 k^3+41 k^2+16 k+1\right)}{(k+1)^2 \left(2 k^8+60 k^7+157 k^6+282 k^4+157 k^2+60 k+2\right)}$$
Trying for a few integer values of $k$, some results $$\left( \begin{array}{ccc} k & \text{estimation} & \text{solution} \\ 2 & 0.419652 & 0.419618 \\ 3 & 0.454355 & 0.454092 \\ 4 & 0.431354 & 0.430701 \\ 5 & 0.400189 & 0.399069 \\ 6 & 0.370539 & 0.368937 \\ 7 & 0.344242 & 0.342186 \\ 8 & 0.321299 & 0.318836 \\ 9 & 0.301296 & 0.298480 \\ 10 & 0.283772 & 0.280656 \\ 20 & 0.182806 & 0.178793 \\ 30 & 0.137075 & 0.133726 \\ 40 & 0.110350 & 0.107907 \\ 50 & 0.092613 & 0.091004\\ 60 & 0.079908 & 0.079001 \\ 70 & 0.070327 & 0.069998 \\ 80 & 0.062830 & 0.062972 \\ 90 & 0.056796 & 0.057322 \\ 100 & 0.051831 & 0.052672 \\ 200 & 0.027734 & 0.029871 \\ 300 & 0.018954 & 0.021282 \\ 400 & 0.014400 & 0.016688 \\ 500 & 0.011612 & 0.013800 \\ 600 & 0.009728 & 0.011806 \\ 700 & 0.008371 & 0.010341 \\ 800 & 0.007345 & 0.009216 \\ 900 & 0.006544 & 0.008323 \\ 1000 & 0.005901 & 0.007600 \end{array} \right)$$ For sure, when $k$ is large, the estimate is poorer and poorer. Let us try for $k=1234.56789$. Newton iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.00479442 \\ 1 & 0.00933283 \\ 2 & 0.00854256 \\ 3 & 0.00778491 \\ 4 & 0.00710836 \\ 5 & 0.00660608 \\ 6 & 0.00636824 \\ 7 & 0.00632569 \\ 8 & 0.00632453 \end{array} \right)$$
Edit
What could be interesting is that we can solve the equation for $k$. It is $$k=\frac{1}{1+e^{-x}}-\frac{W\left(-\frac{ x}{e^x+1}t\right)}{x}\qquad \text{with}\qquad t=e^{\frac{e^x x}{e^x+1}+x}$$ where $W(.)$ is Lambert function. The solution is real as long as $x < 0.455$ and it is valid for any $k$.
So, $x$ is "just" the inverse function of the monster.