How would you find the inverse of a function that is both linear and logarithmic?
Take this for example: $f\left(x\right)=ln\left(x\right)+x+1$
Writing it as $y=ln\left(x\right)+x+1$ won't work, at the end after some algebra I get
$e^{\left(y-1\right)} = xe^x$
Is there a mistake in the calculation or is this not the way I am supposed to find the inverse?
You are asking for closed-form inverses.
A function has an inverse (inverse function) iff it is bijective. Each surjective function can be decomposed into bijective functions (bijective restrictions) by decomposing its domain. The inverses of the bijective restrictions are the partial inverses of the function.
Your function is bijective i.a. for $x>0$.
1.)
Let's discuss functions $f\colon D\ \text{non-discrete}\ \subseteq\mathbb{C}\to\mathbb{C},z\to f(z)=A(z,\ln(z))$ or $f(z)=A(z,e^z)$, where $A$ is an algebraic function.
These functions are Elementary functions. The Elementary functions are generated from their complex argument by applying finite numbers of $\exp$, $\ln$ and/or algebraic functions.
1. a)
With help of Ritt's theorem on the existence of elementary inverses of elementary functions ([Ritt 1925]), one can deduce that a function $f$ as specified above cannot have elementary partial inverses over non-discrete domains.
1. b)
The question of existence of closed-form inverses of functions is different from the question of closed-form solutions of equations. But both questions are interrelated.
Let $F$ an elementary function $D_F\ \text{non-discrete}\ \subseteq\mathbb{C}\to\mathbb{C}$ and let $\alpha$ a variable that represents an elementary number or an algebraic number. If the equation $F(z)=\alpha$ has only a finite number of solutions $z$ that are elementary numbers, the function $F$ cannot have elementary partial inverses over non-discrete domains.
From the theorems in [Lin 1983] and [Chow 1999] follows, that equations $f(z)=\alpha$ ($f$ as specified above, $\alpha$ a variable that represents an algebraic number), where $A$ is a polynomial irreducible over the algebraic numbers, don't have solutions except $0$ that are elementary numbers.
2.)
Your function $f$ is not defined for $x=0$, we have to set $x\neq 0$ therefore.
Your result $e^{y-1}=xe^x$ is correct. But consider point 1.) above: the equation cannot have elementary solutions.
But the problem can be solved in terms of Lambert W.
Lambert W is the inverse relation of the function $x\mapsto xe^x$. We have therefore:
$$e^{y-1}=xe^x$$
$$W_k(e^{y-1})=W_k(xe^x)\ \ \ \ (k\in\mathbb{Z}).$$
Now apply the appropriate branches of Lambert W.
$$W_k(e^{y-1})=x$$
Analogous results can be applied also to functions $f$ with $f(x)=a\ln(x)+bx+c$, wherein $a,b,c\in\mathbb{C}$.
$\ $
[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448
[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50
[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90