Confusing result obtained taking second derivative of ye^y

Question

I was doing my calculus homework, and one of the questions asked for the first and second derivative of $ye^y=x$, I did the computations and arrived at $-(x+1)^{-2}$, which was a lot neater and simpler than I expected, I am particularly perplexed by the fact that this expression doesn't have a $y$ in it, while my first derivative $\bigg(\frac{y}{yx+y}\bigg)$ did.

Is my math correct, and if it is, is there any way this result could be explained non-mathematically, in such a way that I might be able to understand why this would be true?

My math:

$$ye^y=x \rightarrow ln(y) + y = ln(x) \rightarrow \frac{d}{dx}\big[ln(y)+y\big]=\frac{d}{dx}\big[ln(x)\big]\rightarrow\frac{dy}{dx}=\frac{y}{xy+y}$$

for the first derivative, and

$$\frac{d^2y}{dx^2}=\frac{y'(xy+y)-(xy+y)'y}{(xy+y)^2}=\frac{\frac{y}{xy+y}(xy+y)-y[y'(x+1)+(x+1)'y]}{y^2(x+1)^2}=\frac{y-y[\frac{y}{xy+y}(x+1)+y]}{y^2(x+1)^2}=\frac{y-y(1+y)}{y^2(x+1)^2}=\frac{y-y-y^2}{y^2(x+1)^2}=-\frac{1}{(x+1)^2}=-(x+1)^{-2} $$

for the second dericative.

Answer 1

In that step when you first took the derivative of $y$ you should get $\frac{dy}{dx}= \frac{y}{(y+1)x}$, right?

Answer 2

I don't understand what you did, I (and Wolfram Alpha) would do this:

$${d\over dx}\left(ye^y\right) = {d\over dx}\left(x\right)$$

$$\iff {d\over dx}\left(ye^y\right) = 1$$

$$\iff y'e^{y} + y'e^{y}y = 1$$

$$\iff y'\left(e^{y} + e^{y}y\right) = 1$$

$$\iff y'(x) = {1 \over e^{y} + e^{y}y}$$ $$\iff y'(x) = {1 \over e^{y}(1 + y)}$$