If you know Taylor expansion, you know that
$$\tan x = x + \frac{x^3}{3}+ \mathcal{O}(x^5)$$
where the big-Oh denotes a term which scales like $x^5$ for $x\to 0$.
Thus, $$\frac{\tan x}{x} = 1 + \frac{x^2}{3} + \mathcal{O}(x^4).$$
The expansion of the logarithm around $1$ reads
$$ \ln (1+y) = y + \mathcal{O}(y^2).$$
Letting $1+y=\tan x/x =1+ x^2/3 + \mathcal{O}(x^4)$, we obtain
$$ \ln \left(\frac{\tan x}x \right) = \frac{x^2}3 + \mathcal{O}(x^4).$$
Now,
$$ \frac1 {x^2} \ln \left(\frac{\tan x}x \right) = \frac13 + \mathcal{O}(x^2).$$
And thus $$\lim_{x\to 0} \frac1 {x^2} \ln \left(\frac{\tan x}x \right) = \frac13.$$
With that you can easily show that
$$\lim_{x\to 0} \left(\frac{\tan x}x\right) ^{x^{-2}} = \sqrt[3]{e}.$$
Since it looks messy to put my efforts in the question body, I decided to write an answer to explain what I have shown and where I got stuck, in an attempt to get an answer. :)
The overall look:
I tried to show first that $g$ is differentiable, then that $g(x,y)$ is increasing in $x.$ After this, I tried to find the limit $\lim_{y\rightarrow\infty}g(x,y),$ but to no avail. Finally, if I can find the above limit, then I can also compute $\lim_{y\rightarrow\infty}\frac{\partial}{\partial x}g(x,y).$
Differentiability of $g$:
Define $f(x,y,z)=\ln(\frac{1+z}{1-z})+2y\tan^{-1}(yz)-2(y^2+1)x.$
Then, for any pair $(x,y)$ with $x\in\mathbb R, y\gt0,$ we find that $\lim_{z\rightarrow1}f(x,y,z)=\infty,$ and $\lim_{z\rightarrow-1}f(x,y,z)=-\infty.$ Hence, by the mean-value theorem, there is $z_0$ such that $-1\lt z_0\lt1$ and $f(x,y,z_0)=0.$ Thus we can apply the implicit-function theorem, and deduce that there is a nbd $U$ of $(x,y),$ and a differentiable function $g_U:U\rightarrow \mathbb R$ such that $g_U$ is the unique solution to the functional equation on $U.$ This shows that $g\mid_U=g_U,$ and hence $g$ is differentiable everywhere, i.e. $g$ is differentiable.
Monotonicity of $g$ in $x$:
By the implicit-function theorem, we find that $$\frac{\partial}{\partial x}g(x,y)=\frac{2(y^2+1)}{\frac{1}{1+g(x,y)}-\frac{1}{1-g(x,y)}+\frac{2y^2}{1+(yg(x,y))^2}}\gt0.$$
This shows that $g(x,y)$ is increasing in $x.$
On the limit $\lim_{y\rightarrow\infty}g(x,y)$ :
I rewrite the equation as:
$$2y(\tan^{-1}(yg(x,y))-xy)=2x-\ln(\frac{1+g(x,y)}{1-g(x,y)})$$
$$\tan^{-1}(yg(x,y))=xy+\frac{x}{y}-\frac{\ln(\frac{1+g(x,y)}{1-g(x,y)})}{2y}$$
$$g(x,y)=\frac{tan(xy+\frac{x}{y}-\frac{\ln(\frac{1+g(x,y)}{1-g(x,y)})}{2y})}{y}.$$
But I still cannot infer what the limit should be from this.
The Second Limit
is a direct consequence of the implicit-function theorem and the above limit.
This is all I know about the question, in the dearth of a piece to complete the puzzle.
Hope this helps clarify my question.
Best Answer
As you say, if there is such an $f$, then $f'(x) > f(0) > 0$ for all $x$. We can now use that lower bound on the derivative to show that $f$ must be negative for sufficiently large, negative $x$:
Let $x < 0$. Then by the MVT, there exists a $c \in (x, 0)$ such that $$\frac{f(0) - f(x)}{-x} = f'(c) > f(0)$$
Hence $f(0) - f(x) > f(0)(-x) \ \ \ $ or $ \ \ \ \ f(x) - f(0) < f(0)x$.
Thus for $x < -1$,
$$f(x) < (x+1)f(0) < 0$$ contradicting the hypothesis that $f(x)$ is always positive.