How do I find the derivative of $x$ with respect to $y$ when $y = e^{-x^2}$? When I have $y = e^{-x^2}$ with $-1 \leq x \leq 1$, I want to find the derivative of $x$ with respect to $y$. Can I take the log of both sides?
$$\ln y = -x^2$$
From here, can I say that $-x^2$ is always = $x^2$? If so, I get:
$$\ln y = x^2 \implies \sqrt{\ln y} = x$$
What's the derivative?
$$\frac{dx}{dy} = \frac{1}{2} \left(\ln y^{\frac{-1}{2}} \right) \cdot \frac{1}{y} = \frac{1}{2y \sqrt{\ln y}}$$
Is that right?
 A: $$
y=e^{-x^2}\implies\ln{y}=-x^2\implies x=\pm\sqrt{-\ln{y}},\ (0\lt y\le 1)
$$
This means that for $x$ as a function of $y$, you get two separate functions that you have to differentiate separately because those are literally two different curves (the right curve and the left curve which are symmetric with respect to the y-axis):
$$
\left(\sqrt{-\ln{y}}\right)'=-\frac{1}{2\ln{y}}\left(-\ln{y}\right)'=\frac{1}{2y\ln{y}}\\
\left(-\sqrt{-\ln{y}}\right)'=-\left(-\frac{1}{2\ln{y}}\left(-\ln{y}\right)'\right)=-\frac{1}{2y\ln{y}}
$$
A: So, context - OP originally made a question that implied $dy/dx$ to all of the answerers and not $dx/dy$. This question is now about the latter. I addressed most of the fatal flaws of OP's method, which were copied and pasted from the original question, in my original post here. Most of the same flaws render the method invalid again here (since they're just generic flaws and not tied to the method or question). So I'm not going to reiterate them here, it's not particularly necessary.
This question is the third in a series by OP, the original of which was this, a question about finding the surface area of the solid of revolution formed by $y$ for $-1 \leq x \leq 1$. I note this because it will be relevant.

So, to begin -  yes, the best idea here is to solve for $x$ first. When possible, in a scenario like this, it makes life simpler, and it works here. 
Here, to solve for $x$, we take the natural logarithm of both sides, multiply by negative $1$, and then take the square root:
$$y = e^{-x^2} \implies \ln(y) = -x^2 \implies x^2 = -\ln(y) \implies x = \pm \sqrt{-\ln(y)}$$
A paranoia check: does this make sense? We have $-1 \le x \le 1$, and $y$ has values ranging from $0$ to $1$ for any real $x$ (you can see this if you check a graph or know anything about the normal distribution). Since $0 < y \le 1$, $-\infty < \ln(y) \le 0$, which means that the radicand is always nonnegative. Good, everything makes sense.
Sort of. Strictly speaking this means $x$ can be anything from positive to minus infinity, what is the relevance of the $x \in (-1,1)$ bit? That is a restriction of the original problem, which is concerned with the surface area of a solid of revolution. In other words, it's not a particularly important restriction in the sole issue of finding $dx/dy$ so we will ignore it until it becomes relevant again. But rest assured, strictly speaking, this answer "makes sense" (we don't have any imaginary numbers popping up)
So, we have,
$$x = \pm \sqrt{-\ln(y)} = \pm (-\ln(y))^{1/2}$$
If you differentiate this with respect to $y$, we have a standard chain rule calculation, which yields
$$x' = \pm \frac 1 2 \cdot(-\ln(y))^{-1/2} \cdot \frac 1 y = \pm \frac{1}{2y\cdot \sqrt{-\ln(y)}}$$
So ironically as it happens, up to the sign of the square root and logarithm, you were kinda right all along. Albeit partly because of the wrong reasons. But oh well, now you know, hopefully. :)
Okay but what's up with the $\pm$ sign? That's because of how $x$ is sometimes negative and sometimes positive, as noted by Jack in the answers to your original question. You will use the positive $x$ for $x\in [0,1]$ and negative for $x \in [-1,0)$.
EDIT: Of course, as noted by Minus One-Twelfth in the comments, you're going to be squaring $dx/dy$ anyways so the $\pm$ isn't of huge consequence regardless.
A: I tend to look at this problem in terms of what is asked, and solving for $x$ is neither asked nor required. 
I suggest you use implicit differentiation.
$$y=e^{-x^2}$$
$$dy =  -2x e^{-x^2}dx.$$
$$ \frac{dx}{dy} = \frac{1}{-2x e^{-x^2}}$$
$$ \frac{dx}{dy} = -\frac{1}{2xy}.$$
