Say $f(x)\in\mathbb{R}$ is defined by $$f(x)=f(x)^{\frac{x-1}{x}}+x$$ For example, $f(95)$ is around $1306$ since $1306^{94/95}+95\approx1306$. A friend found this and was trying to find its growth rate. Just something like $f(x)\sim c\cdot x^k$.
Just from eyeballing the growth rate for small $x$, it looks like $k$ is a bit under $2$ but I couldn't find any way to prove this. I did notice that $$``f'(x)"\approx f(x+1) - f(x)=f(x+1)^{\frac{x}{x+1}}-f(x)^{\frac{x}{x+1}\cdot\frac{x^2-1}{x^2}}+1$$ which makes me want to say $f$'s derivative is simply $1$ since $\frac{x^2-1}{x^2}$ tends to $1$. But this isn't rigorous or even well defined. Part of me feels like the Lambert-W function needs to get involve since directly substituting $f(x)=c\cdot x^k$ leaves us with $x$ to the power of $x$ (or at least, expressions involving $x$).
EDIT: forgot to add, $f(x)$ can be equivalently defined with $$f(x)^{-1}=\Big(1-\frac{x}{f(x)}\Big)^x$$ which I didn't find to be particularly helpful. The equation looks like a weird variant of $e$'s limit definition though and is interesting.
Collecting terms in $f$ on the LHS, we have $f\cdot\left(1-f^{-{1}/{x}}\right)=f\cdot\left[1-e^{-\ln f/x}\right]=x$. It would be reasonable to assume, from graphing or trial computations, that if $x\to\infty$, $\ln f$ might at least be small enough that $\frac{\ln f}{x}\to0$ (i.e., $f$ is subexponential). In that case, we could expand the exponential as a power series such that $f\cdot\frac{\ln f}{x}\approx x$.
That would solve for $f(x)\approx \exp W(x^2)= \exp\left[L_1(x^2)-L_2(x^2)+\ldots\right]$, where $W$ is the Lambert W, which has a well known series in $L_1(x)=\ln x$ and $L_2(x)=\ln \ln x$. Hence, $f(x)\approx \frac{x^{2}}{2\ln x}$, which seems close to its order.
Edit: To conclude this solution as a satisfactory proof, we could prove that the supposed order does satisfy bounds involving $f$. Using the given definition of $f$, define a new explicit, univariate function with constant $c>0$
$$g(x)=\left(c\frac{x^{2}}{\ln x}\right)\left[1-\left(c\frac{x^{2}}{\ln x}\right)^{-\frac{1}{x}}\right]-x$$
Hence, the zeros of $g(x)$ are the $x$ that satisfy $c\frac{x^{2}}{\ln x}=f(x)$, so those zeros' behavior may describe how $f(x)$ and $\frac{x^{2}}{\ln x}$ relate, for a given $c$.
From the Taylor series identity $e^x\ge 1+x$, we have (like in the earlier, loose argument),
$$g(x)\le \left(c\frac{x^{2}}{\ln x}\right)\left[\frac{1}{x}\ln\left(c\frac{x^{2}}{\ln x}\right)\right]-x$$
Some simple algebra shows that the RHS of this inequality is zero iff $cx^{2-{1}/c}=\ln x$. But if $c=1/2$, the LHS of this is a constant and hence only admits one zero (at $x=1.649$). Otherwise, for sufficiently large $x$, we have $g(x)<0$, so $c\frac{x^{2}}{\ln x}=f(x)$ will have no solutions and therefore $f(x)>\frac12\cdot\frac{x^{2}}{\ln x}$ as the two do not intersect.
Conversely, we have the identity that $e^x\le \frac1{1-x}$ for all $x<1$. (To ensure that this is valid to use, we now also assume $c>1/2,x>1$ so that $c\frac{x^{2}}{\ln x}>1$ and $-\frac{1}{x}\ln\left(c\frac{x^{2}}{\ln x}\right)$ is strictly negative.) Hence,
$$g(x)\ge \frac{cx^{2}\ln\left(c\frac{x^{2}}{\ln x}\right)}{\ln x\left[x+\ln\left(c\frac{x^{2}}{\ln x}\right)\right]}-x$$
For $c=1$, the RHS of this is zero iff $x\left(x-\ln\left(x\right)\right)\ln\left(\frac{x^{2}}{\ln\left(x\right)}\right)-x^{2}\ln\left(x\right)=0$, however that expression is strictly positive (I will leave this unproven, but it should be doable, if slightly tedious). Hence, in this case $g$ also has no zeros, so $f(x)<1\cdot\frac{x^{2}}{\ln x}$.
Therefore, for sufficiently large $x$, $f(x)$ is bounded between $\frac12\cdot\frac{x^{2}}{\ln x}$ and $1\cdot\frac{x^{2}}{\ln x}$, so it has order $\Theta\left(\frac{x^{2}}{\ln x}\right)$.