$a\\$ - edge length at the base of the pyramid.
$b\\$ - length of the side edges of the pyramid.
$V\\$ = $const\\$.
I minimize: $3a+3b->min\\$
$V=\frac{1}{3}Sh\\$
$h=\sqrt{b^2-\frac{1}{3}a^2}\\$
$S=\frac{\sqrt{3}}{4}a^2\\$
$V=\frac{1}{3}*\frac{\sqrt{3}}{4}a^2*\sqrt{b^2-\frac{1}{3}a^2}\\$ - from here I expressed the parameter $b\\$ and substituted in $3a+3b->min\\$
$b=\sqrt\frac{V^2+\frac{3a^6}{432}}{\frac{3a^4}{144}}\\$
After substitution, I found the derivative and equated it to zero, but the derivative was too complex, it seems to me that I am doing something wrong.
Derivative: $3+\frac{a^6-288V^2}{a^5*\sqrt{\frac{48V^2}{a^4}+\frac{a^2}{3}}}=0\\$
How to express $a\\$ from here?
As brought up in the comments, we have that
$$ b=\sqrt{\frac{144{V^2}}{3a^4}+\frac{1}{3}a^2}. $$
You wish to minimize the function
$$ f(a)=a+\sqrt{\frac{144{V^2}}{3a^4}+\frac{1}{3}a^2} $$
by equating the derivative to zero (note that I dropped the scaling multiplication by $3$, since it does not affect the minimizer). Introducing the notation $k=144V^2/3$ and $m=1/3$, the derivative of $f$ is
$$ f'(a)=1-\frac{\big(\frac{2k}{a^5}-ma\big)}{\sqrt{\frac{k}{a^4}+ma^2}} $$
Equate this to zero, and get
$$ \sqrt{\frac{k}{a^4}+ma^2}=\frac{2k}{a^5}-ma $$
Multiply both sides by $a^5$:
$$ \sqrt{ka^6+ma^{12}}=2k-ma^6. $$
Introduce the notation $x\equiv{a^6}$, and this becomes
$$ \sqrt{mx^2+kx}=2k-mx. $$
Square both sides (at this point we need to be careful about choosing the right signs):
$$ m(1-m)x^2+k(1+4m)x-4k^2=0. $$
Solve with the quadratic formula (choose the proper sign), and substitute $x=a^6$ to get the final answer.