Antiderivative of $-\frac{1}{e^{3x}}$ Solving, $\dfrac{-1}{e^{3x}}$ has different values by integrating using $u$ substitution if $3x$ is on the top or bottom, can someone see what I'm doing wrong? The top answer is negative while the bottom answer is positive.

My work:
$$\mathrm{d}x+e^{3x}\mathrm{d}y=0 \Longrightarrow\int\mathrm{d}y=\int-\frac{\mathrm{d}x}{e^{3x}}\Longrightarrow y=-\int\frac{1}{e^u}\frac{\mathrm{d}u}{3}\Longrightarrow y=-\frac13\times\frac{1}{e^{3x}}$$
With $u=3x$ and $\mathrm{d}u=3\mathrm{d}x$ or $\mathrm{d}x=\frac{\mathrm{d}u}3$.
On the other hand:
$$y=-\int e^{-3x}\mathrm{d}x\Longrightarrow -\int e^u\frac{\mathrm{d}u}{-3}\Longrightarrow \frac13e^{-3x}$$
With $u=-3x$ and $\mathrm{d}u=-3\mathrm{d}x$ or $\mathrm{d}x=\frac{\mathrm{d}u}{-3}$.

 A: (Significant edits thanks to Jean-Claude's comment below.)
The major error is in the first one: $\int\frac{1}{e^u}du$ is not $\frac{1}{e^u}$, it's $-\frac{1}{e^u}$. $\int e^udu$ is the straightforward one - in order to use it for $\frac{1}{e^u}$, you'd have to use another $u$-substitution, like the one you did in the second version. 
A: Your second computation is correct.
In the first one, there is a mistake in:
$$ y=-\int\frac{1}{e^u}\frac{\mathrm{d}u}{3}\Longrightarrow y=-\frac13\times\frac{1}{e^{3x}}$$
It should be:
$$y=-\int\frac{1}{e^u}\frac{\mathrm{d}u}{3}=-\frac13\int \frac{\mathrm{d}u}{e^u}=-\frac13\int e^{-u}\mathrm{d}u=\frac{1}{3}e^{-u}=\frac{1}{3}e^{-3x}$$
The antiderivative of $e^u$ is $e^u$, but the antiderivative of $e^{-u}$ is $-e^{-u}$.
Notice that generally you can simplify such change of variable $u=ax$ by remembering that if $F(x)$ is an antiderivative of $f(x)$, then $\frac1aF(ax)$ is an antiderivative of $f(ax)$.

Also, it should be obvious, but keep in mind that all these indefinite integrals should have a constant term added in the final result. Here it does not harm, but sometimes you end up with different nontrivial expressions that happen to differ by a nonzero constant. A usual example is
$$\int\frac{1}{\cosh x}\mathrm dx=2\arctan(e^x)+C$$
$$\int\frac{1}{\cosh x}\mathrm dx=2\arctan(\tanh\frac x2)+C$$
But 
$$2\arctan(e^x)=2\arctan(\tanh\frac x2)+\frac{\pi}4$$
