As you suspect, using numerical methods will solve the problem easily but we shall not use any of them and we shall try to approximate the solution of the transcendental equation which, by definition, cannot show any analytical expression.
Considering the first derivative
$$y'(t)= \frac{9}{2}e^{-t/2} + 8 \cos(t) - 4 \sin( t) $$ and using your observation from the plot, let us consider Taylor series built around $t=a$. They are $$e^x=e^{-a}-e^{-a} (x-a)+\frac{1}{2} e^{-a} (x-a)^2+O\left((x-a)^3\right)$$ $$\sin (x)=\sin (a)+(x-a) \cos (a)-\frac{1}{2} (x-a)^2 \sin (a)+O\left((x-a)^3\right)$$
$$\cos(x)=\cos (a)-(x-a) \sin (a)-\frac{1}{2} (x-a)^2 \cos (a)+O\left((x-a)^3\right)$$ Applying them (using for simplicity $a=\frac \pi 2$), we then have
$$y'(t)=\left(\frac{9 e^{-\pi /4}}{10}-\frac{4}{5}\right)-\left(\frac{8}{5}+\frac{9
e^{-\pi /4}}{20}\right) \left(t-\frac{\pi }{2}\right)+\left(\frac{2}{5}+\frac{9
e^{-\pi /4}}{80}\right) \left(t-\frac{\pi }{2}\right)^2+O\left(\left(t-\frac{\pi
}{2}\right)^3\right)$$ which then reduces to a quadratic equation in $\left(t-\frac{\pi }{2}\right)$.
Using the classical formula, we then arrive to $$y'(t)=0 \implies t=\frac{\pi }{2}+\frac{2 \left(9+32 e^{\pi /4}-\sqrt{3 \left(-27+48 e^{\pi /4}+512 e^{\pi
/2}\right)}\right)}{9+32 e^{\pi /4}}\approx 1.36548$$ while the exact solution would be $t\approx 1.36431$ which seems to be quite good.
Using the result of the approximation, the maximum would be $\approx 0.820081$ while the exact solution would be $\approx 0.820082$.
Edit
We could have done better using $a=\frac{5 \pi }{12}$ (we know the values of the trigonometric functions for this angle). Doing the same as above, we should have
$$y'(t)=\left(-3 \sqrt{2}+\sqrt{6}+\frac{9}{2} e^{-5 \pi /24}\right)-\left(\sqrt{2}+3
\sqrt{6}+\frac{9}{4} e^{-5 \pi /24}\right) \left(t-\frac{5 \pi
}{12}\right)+\left(\sqrt{6-3 \sqrt{3}}+\frac{9}{16} e^{-5 \pi /24}\right)
\left(t-\frac{5 \pi }{12}\right)^2+O\left(\left(t-\frac{5 \pi
}{12}\right)^3\right)$$ Ignoring the quadratic term, the zero would be given as $t\approx 1.36392$. Using the quadratic term, $t\approx 1.36429$.
Final Answer: (to summarize)
In general, if we have a function of the form f(x)=|g(x)|, then you can note that f(x)⩾0 for all x in the domain. Moreover, if there are points at which g(x) is zero, then they must be global minima for the function f. (USEFUL POINT TO REMEMBER)
If you know a function f:U→R for U open has a singularity at a point then limx→p|f(x)|=∞, This means that it is for certain that f has either no global min or no global max.
If yet there is any other critical point, just use the fundamental definition that $f\left(c^{-}\right)>f\left(c\right)\ and\ f\left(c^{+}\right)>f\left(c\right)$ for minima and like that we can do for maxima. This can be used in a piece-wise function too at the point where the definition changes, to find if it is an extrema or not.
It is not a hard and fast method and depending upon the questions, things can change, nothing can be said exactly. Best way to solve these questions is the graphical approach wherever applicable.
Best Answer
For $x>0$,
As you found $$f'(x)=\frac{2-\ln(x)}{2x\sqrt{x}}$$
the deniminator is $>0$, the sign of $ f'(x) $ is given by the sign of the numerator $ 2-\ln(x) $.
$$f'(x)=0\iff \ln(x)=2$$ $$\iff x=e^2$$
$$f'(x)>0\iff 2>\ln(x)$$ $$\iff 0<x<e^2$$
thus $ f $ is increasing at $ (0,e^2] $ and decreasing at $ [e^2,+\infty)$.
$ f $ attains its maximum at $ x=e^2$. on the other hand $$\lim_{x\to 0^+}f(x)=-\infty$$ and $$\lim_{x\to+\infty}f(x)=0$$ So, it has no minimum.