Note that $2\log_b u = \log_b u^2$ for any $b$ and $u$, so that we can rewrite this as
$$
\log_6 \left(x+2\sqrt[4]{x^3}+\sqrt{x}\right) = \log_4 x
$$
From here, the easiest thing is to guess, and observe that for the argument of the left-side log comes out nicely if $x$ is a fourth power. For instance, we might guess $x = 2^4 = 16$, which yields $x+2\sqrt[4]{x^3}+\sqrt{x} = 16+2(8)+4 = 36$, and indeed
$$
\log_6 36 = 2 = \log_4 16
$$
ETA: That this is the only solution for $x > 0$ can be seen if we write (following GEdgar's comments to the original post)
$$
\frac{\log_4 \left(x+2\sqrt[4]{x^3}+\sqrt{x}\right)}{\log_4 6} = \log_4 x
$$
$$
\frac{\log_4 \left(x+2\sqrt[4]{x^3}+\sqrt{x}\right)}{\log_4 x} = \log_4 6
$$
or equivalently
\begin{align}
\log_4 6 & = \log_x \left(x+2\sqrt[4]{x^3}+\sqrt{x}\right) \\
& = 1 + \log_x \left(1+\frac{2}{\sqrt[4]{x}}+\frac{1}{\sqrt{x}}\right) \\
& = 1 + 2\log_x \left(1+\frac{1}{\sqrt[4]{x}}\right)
\end{align}
Note that the argument of the $\log_x$ is a decreasing function of $x$, which means that the entire right-hand side is a decreasing function of $x$ for $x > 1$, which in turn means that the solution $x = 16$ is the only solution for $x > 1$.
For $0 < x < 1$, the argument of the $\log_x$ is greater than $1$, but its base $x$ is less than $1$, so the log is negative, and the right-hand side is less than $1$, whereas the left-hand side is greater than $1$, so there are no solutions in that range, either.
Finally, $x = 1$ is not a solution of the original equation, by inspection.