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Use the substitution rule with $u=\sqrt{x}$, for which $du=dx/(2\sqrt{x})$ on $x>0$. The integral becomes $\int e^{u}du$, giving the antiderivative

For $x>0$, the denominator is positive, so the function graph has the same sign as $\sin x$. Its crossings of the horizontal axis areEach crossing is genuine because the derivative there is $(-1{)}^n/(n\pi {)}^2\ne 0$. Near zero, $\sin x\sim x$, so $y\sim 1/x\to +\infty$; the origin is excluded and is not a crossing. The oscillations alternate sign and decay towards zero as $x\to \infty$, with absolute value at most $1/x^2$.

The product rule givesFor $x>0$, the derivative is positive before $3$ and negative after $3$. Also $y(0)=0$ and $x^3{e}^{-x}\to 0$ as $x\to \infty$. Thus the global maximum on the requested range is

The original PDF puts both $x+y$ in the exponent in the second equation. Since the exponential function $3^t$ is injective, $3^{x+y}=1$ gives $x+y=0$. Put $u=3^x>0$, so $3^y=1/u$. The first equation then givesThus $u=1$, so $x=0$ and $y=-x=0$. The unique real solution of the simultaneous equations is

The domain excludes $x=0$. Rearranging the inequality givesA sign check on the intervals separated by $-1$ and $0$ shows that the ratio is nonnegative when $x\le -1$ or $x>0$. The point $-1$ is allowed because it makes the numerator zero. Therefore

Completing the square under the square root gives $\sqrt{(x-1{)}^2}=|x-1|$. Thus the equation is $|x-1|=2$, and the definition of absolute value gives $x-1=\pm 2$. Hence