Does $e^{tx} + x = t - 1$ yield a solution/s to $\frac{dx}{dt} = \frac{e^{-tx} - x}{e^{-tx} + t}$? 
Problem: Use Existence Theorem to determine if $x(t)$ implicitly defined by
$$e^{tx(t)} + x(t) = t - 1 \tag{*}$$
yield a solution/s to
$$\frac{dx}{dt} = \frac{e^{-tx} - x}{e^{-tx} + t} \left( = \frac{1 - xe^{tx}}{1 + te^{tx}} \right) \tag{**}$$
?



Definition: Let $x = x(t)$. A solution of the first order ODE $x' = f(t,x)$, where $f$ is defined on some domain $D \subseteq \mathbb R^2$ s.t. $D$ is open and connected, is a differentiable function $\varphi$ on some interval $I \subseteq \mathbb R$ s.t.

*

*$(t, \varphi(t)) \in D \ \forall t \in I$


*$\varphi'(t) = f(t, \varphi(t)) \ \forall t \in I$



Existence Theorem: If $f$ is continuous on a domain $D \subseteq \mathbb R^2$ and $(\tau, \xi) \in D$, then $\exists$ a solution $\varphi$ of $x' = f(t,x)$ defined on some interval $I$ s.t. $\tau \in I$ and $\varphi(\tau) = \xi$


What I tried:
The denominator vanishes when $x=0$ and $t=-1$ so I guess we can't say $D = \mathbb R^2$. How about
$$D = \{(t,x) | (t,x) \ne (-1,0) \}$$
?
Is that open and connected? We didn't have any topology courses, and our real analysis only covered $\mathbb R$.
If that's right, $f$ I guess is continuous on $D$.
Choose some point $(\tau, \xi) \in D$:
$(\tau, \xi) = (1,0)$
So $\exists$ a solution $\varphi$ of $x' = f$ defined on an interval $I \subseteq \mathbb R$ s.t. $1 \in I$ and $\varphi(1) = 0$.
Is that right?
 A: Just implicitly compute the derivative, we see
$$e^{tx(t)} + x =t-1  \implies  (x + t x' ) e^{tx} + x' =1 \implies x' ( 1+ te^{tx})= 1- xe^{tx} \implies x' = \frac{1 -xe^{tx} }{1 +te^{tx}} $$
Since the denominator is bad along $1+te^{tx}=0$, this is line we avoid for $x(t)$. By the implicit function theorem, we may always find $x(t)$ everywhere else. 
A: $$x' = \frac{1 - xe^{tx} }{1 + te^{tx}}$$
$$x' + te^{tx}x'+xe^{tx} -1=0$$
$$\left(x+e^{tx}-t \right)'=0$$
$$x+e^{tx}-t = c$$
This is the inplicit form of all the solutions. 
The proposed solution $e^{tx}+x=t-1$ is one of them (particular case $c=-1$ ).
NOTE :
An explicit form of the solutions can be derrived, thanks to the Lambert's W function.
$$e^{tx}=-x+t+c$$
$$ (-x+t+c)e^{-tx}=1$$
$$ (-tx+t^2+ct)e^{-tx}=t$$
$$ (-tx+t^2+ct)e^{-tx+t^2-ct}=te^{t^2-ct}$$
with $W= (-tx+t^2+ct)$ and $X=te^{t^2-ct}$ then  $W e^W=X$ where $W=W(X)$
$$-tx+t^2+ct=W\left(te^{t^2-ct}\right)$$
$$x=t+c-\frac{1}{t}W\left(te^{t^2-ct}\right)$$
According to the properties of the Lambert W function, $W(X)$ returns one value if $X>0$, or two values if $-\frac{1}{e}<X<0$ and no real value if $X<-\frac{1}{e}$
