How do I find the derivative of $y = e^{-x^2}$ with respect to $y$? I want to find the derivative of $y = e^{-x^2}$ with $-1 \leq x \leq 1$ 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: The most straightforward way to differentiate
$y = e^{-x^2} \tag 1$
that I know is via the chain rule, setting
$u = -x^2; \tag 2$
then
$y = e^u, \tag 3$
and so
$y' = \dfrac{dy}{du}\dfrac{du}{dx} = e^u(-2x) = -2xe^{-x^2}. \tag 4$
Some folks like logarithmic differentiation, viz
$\ln y = -x^2, \tag 5$
and so
$\dfrac{y'}{y} = (\ln y)' = -2x, \tag 6$
or
$y' = -2xy = -2xe^{-x^2}, \tag 7$
which as anticipated agrees with (4).
A: Yes, you can do that. It's called logarithmic differentiation. Take logs of both sides and differentiate them with respect to $x$. You don't need to square or take the square root of anything. And there are also no domain issues either: $e^{-x^2}>0$. You can take the log of that without any problem. Here are the steps:
$$\begin{align}
y&=e^{-x^2}\\
\ln{y}&=\ln{e^{-x^2}}\\
\ln{y}&=-x^2\ln{e}\\
\ln{y}&=-x^2\\
\frac{d}{dx}\left(\ln{y}\right)&=\frac{d}{dx}\left(-x^2\right)\\
\frac{1}{y}y'&=-2x\\
y'&=-2yx\\
y'&=-2e^{-x^2}x
\end{align}$$
Thus: $$\left(e^{-x^2}\right)'=-2xe^{-x^2}.$$
A: Personally, I would just use the chain rule. Let $f(x) = -x^2, g(x) = e^x$ for simplicity's sake. Then
$$y = e^{-x^2} = (g \circ f)(x)$$
Thus,
$$y' = (g(f(x))' = g'(f(x)) \cdot f'(x)$$
Since $g'(x) = e^x, f'(x) = -2x$, then
$$y' = e^{-x^2} \cdot (-2x) = -2xe^{-x^2}$$

To address your particular approach, $-x^2 \neq x^2$. This touches on the whole "$-x^2$ is not $(-x)^2$" debacle. Basically, $-x^2$ means $(-1) \cdot x^2$. Notice how this creates issues when you start taking the square root.
Also in your derivative step you wrote $dx/dy$ when you probably meant $dy/dx$.

EDIT: A further issue: $\sqrt{\ln y}$ is not $\ln{y^{1/2}}$. Note that $\ln(y^a) = a \ln(y)$, that meaning that the power has to be inside the logarithm, not applied on the logarithm altogether.
EDIT $2$: I won't really comment on the rest since the $-x^2$ debacle and the rest basically makes the rest of your approach invalid. The broad idea - taking the natural logarithm and using implicit differentiation - is valid, you just applied it wrong as shown in Michael Rybkin's answer. In summary:
$$y = e^{-x^2} \implies \ln y = \ln e^{-x^2} = -x^2 \implies (\ln y)' = \frac{y'}{y} = -2x = (-x^2)$$
By $y'/y = -2x$, then $y' = -2xy$. Substitute in our definition for $y$ to get our answer, same as the one per the chain rule.
