= Solution
Put $\Gamma(z)=I(z)$ for $\Re z>0$. Integration by parts gives $\Gamma(z+1)=z\Gamma(z)$, both proving $\Gamma(n)=(n-1)!$ and defining the meromorphic continuation. A keyhole-contour evaluation of $\int_0^\infty t^{z-1}/(1+t)\,dt$ gives
$$
\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}.
$$
For
$$
\Gamma(z)\Gamma(z+\tfrac12)=e^{g(z)}\Gamma(2z),
$$
the reflection formula at $z$ and $z+\tfrac12$ gives $g(z+1)-g(z)=-2\log2$. Thus $h(z)=g(z)+2z\log2$ is entire and $1$-periodic; poles and zeros have cancelled in the quotient. Its Fourier expansion and the exponential growth bound make it a finite Laurent <polynomial> in $e^{2\pi iz}$; the same growth bound for $e^h$ excludes every nonconstant mode. Hence $h=b$. At $z=1/2$, $e^b=2\sqrt\pi$, yielding the duplication formula.
Set $z=1/6$ and use reflection at $1/3$:
$$
\Gamma(\tfrac16)\Gamma(\tfrac23)=2^{2/3}\sqrt\pi\,\Gamma(\tfrac13),\qquad
\Gamma(\tfrac13)\Gamma(\tfrac23)=\frac{2\pi}{\sqrt3}.
$$
Therefore
$$
\boxed{\Gamma(\tfrac16)=2^{-1/3}\sqrt{\frac3\pi}\,\Gamma(\tfrac13)^2.}
$$
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