System Of Differential Equations!

$(1) \:\ gh * (f' = 2f^2gh + \dfrac{1}{gh})$

$(2) \:\ fh * (g' = fg^2h + \dfrac{4}{fh})$

$(3) \:\ fg * (h' = 3fgh^2 + \dfrac{1}{fg})$

$(fgh)' = f(gh)' + (gh)f' = f(gh' + hg') + f'gh = f'gh + fg'h + fgh' \implies$

$(fgh)' = 6((fgh)^2 + 1)$

Let $y = fgh \implies y' = 6(y^2 + 1) \implies \displaystyle\int \dfrac{y'}{y^2 + 1} dy = 6\displaystyle\int dx$

$\implies \arctan(y) = 6x + C$ and $f(0)g(0)h(0) = 1 = \tan(C) \implies C = \dfrac{\pi}{4} \implies$

$fgh = \tan(6x + \dfrac{\pi}{4})$

Dividing $(1)$ by $f \implies \dfrac{f'}{f} = 2fgh + \dfrac{1}{fgh} = 2\tan(6x + \dfrac{\pi}{4}) + \cot(6x + \dfrac{\pi}{4})$

$\implies \ln(f) = \dfrac{2}{6}\displaystyle\int \tan(6x + \dfrac{\pi}{4}) + \dfrac{1}{6} \displaystyle\int \cot(6x + \dfrac{\pi}{4}) + C^{*} =$

$\ln(K(\dfrac{\sin(6x + \dfrac{\pi}{4})}{\cos^2(6x + \dfrac{\pi}{4})})^{\frac{1}{6}}) \implies$

$f(x) = K(\dfrac{\sin(6x + \dfrac{\pi}{4})}{\cos^2(6x + \dfrac{\pi}{4})})^{\frac{1}{6}}$

$f(0) = 1 \implies 1 = K(\sqrt{2})^{\frac{1}{6}} \implies K = (\dfrac{1}{\sqrt{2}})^{\frac{1}{6}}$

$f(x) = (\dfrac{1}{\sqrt{2}}(\dfrac{\sin(6x + \dfrac{\pi}{4})}{\cos^2(6x + \dfrac{\pi}{4})}))^{\frac{1}{6}}$ and $\boxed{f(\dfrac{\pi}{12}) = 1}$

Similarly for $g(x)$ and $h(x)$ :

Dividing $(2)$ by $g \implies \dfrac{g'}{g} = fgh + \dfrac{4}{fgh} = \tan(6x + \dfrac{\pi}{4}) + 4\cot(6x + \dfrac{\pi}{4}) \implies$

$g(x) = K(\dfrac{\sin^4(6x + \dfrac{\pi}{4})}{\cos(6x + \dfrac{\pi}{4})})^{\frac{1}{6}}$

and $g(0) = 1 \implies K = (2^{\frac{3}{2}})^{\frac{1}{6}} \implies$

$g(x) = (2^{\frac{3}{2}}(\dfrac{\sin^4(6x + \dfrac{\pi}{4})}{\cos(6x + \dfrac{\pi}{4})}))^{\frac{1}{6}}$ and $\boxed{g(\dfrac{\pi}{4}) = 1}$

Dividing $(3)$ by $h \implies \dfrac{h'}{h} = 3fgh + \dfrac{1}{fgh} = 3\tan(6x + \dfrac{\pi}{4}) + \cot(6x + \dfrac{\pi}{4}) \implies$

$h(x) = K(\dfrac{\sin(6x + \dfrac{\pi}{4})}{\cos^3(6x + \dfrac{\pi}{4})})^{\frac{1}{6}}$

and $h(0) = 1 \implies K = (\dfrac{1}{2})^{\frac{1}{6}} \implies$

$h(x) = (\dfrac{1}{2}(\dfrac{\sin(6x + \dfrac{\pi}{4})}{\cos^3(6x + \dfrac{\pi}{4})}))^{\frac{1}{6}}$ and $\boxed{h(\dfrac{\pi}{6}) = 1}$

$\implies f(\dfrac{\pi}{12}) + g(\dfrac{\pi}{4}) + h(\dfrac{\pi}{6}) = \boxed{3}$