Min $f(a,b,c,d)=13abcd$
s.t. $a/3+b^2/5+c^2/4+d^2/6=1$
Let me tell you what I have done.
I took derivative w.r.t. $a,b,c,d$ i.e. $13abcd -\lambda (a/3+b^2/5+c^2/4+d^2/6-1)=0$
$13bcd=\lambda /3$ ----- (1)
$13acd=\lambda 2b/5$ ----- (2)
$13abd = \lambda c/2$ ----- (3)
$13abc= \lambda d/3$ ----- (4)
$a/3+b^2/5+c^2/4+d^2/6=1$
I divided equation (1) and (2), and I did many divisions like that, but I still couldn't solve. Can some one please help me?
Considering $$F=13 a b c d-\lambda \left(\frac{a}{3}+\frac{b^2}{5}+\frac{c^2}{4}+\frac{d^2}{6}-1\right)$$ you properly showed that $$F'_a=13 b c d-\frac{\lambda }{3} \qquad F'_b=13 a c d-\frac{2 b \lambda }{5} \qquad F'_c=13 a b d-\frac{c \lambda }{2}\qquad F'_d=13 a b c-\frac{d \lambda }{3} \qquad F'_\lambda=\frac{a}{3}+\frac{b^2}{5}+\frac{c^2}{4}+\frac{d^2}{6}-1$$ Assuming that $a,b,c,d$ are all $\neq 0$, multiply each of the first four partial derivatives by the missing term and set them equal to $0$.
This should give $$13abcd=\frac{a\lambda }{3}=\frac{2 b^2 \lambda }{5}= \frac{c^2 \lambda }{2}=\frac{d^2 \lambda }{3}=k$$ Now, solve for $a,b,c,d$ and keep the positive solution (this is an assumption for the time being); for sure, we also assume (for the time being) $k>0$. This makes quite many assumptions.
This should give $$a=\frac{3 k}{\lambda }\qquad b=\frac{\sqrt{\frac{5}{2}} \sqrt{k}}{\sqrt{\lambda }}\qquad c=\frac{\sqrt{2} \sqrt{k}}{\sqrt{\lambda }}\qquad d=\frac{\sqrt{3} \sqrt{k}}{\sqrt{\lambda }}$$ Now, replace the above in $F'_\lambda$; this will give $$F'_\lambda=1-\frac{5 k}{2 \lambda }=0\implies \lambda=\frac{5 k}{2}$$ Replacing in $a,b,c,d$, this gives $$a=\frac{6}{5}\qquad b=1\qquad c=\frac{2}{\sqrt{5}} \qquad d=\sqrt{\frac{6}{5}}\implies 13abcd=\frac{156 \sqrt{6}}{25}$$
However, you must consider the other possible solutions since, assuming $k>0$, the solutions are $$a=\frac{3 k}{\lambda }\qquad b_{1,2}=\pm\frac{\sqrt{\frac{5}{2}} \sqrt{k}}{\sqrt{\lambda }}\qquad c_{1,2}=\pm\frac{\sqrt{2} \sqrt{k}}{\sqrt{\lambda }}\qquad d_{1,2}=\pm\frac{\sqrt{3} \sqrt{k}}{\sqrt{\lambda }}$$ In the spirit, you must also condider the case $k<0$.
I suppose that the idea is this kind of stuff.