You can compute the derivative:$$f'(x) = 4^x \ln 4 + 8^x \ln 8 + 9^x \ln 9 - 2^x \ln 2 - 6^x \ln 6 - 12^x \ln 12.$$My instinct would be that a "nice" value of $x$ will probably make $f'(x) = 0$ (probably an integer). So, I would look for $x$ where this expression is $0$, by treating the $\ln 2$ and $\ln 3$ terms separately, hoping both go to $0$ at the same time. Factoring in this way, we get\begin{align}f'(x) &= (2 \cdot 4^x + 3\cdot 8^x - 2^x - 6^x - 2 \cdot 12^x) \ln 2 + (2 \cdot 9^x - 6^x - 12^x) \ln 3 \\&= 2^x(2 \cdot 2^x + 3 \cdot 4^x - 1 - 3^x - 2 \cdot 6^x) \ln 2 + 3^x(2\cdot 3^x - 2^x - 4^x)\ln 3.\end{align}From eyeballing the $\ln 3$ term, $x = 1$ looks plausible, and we can confirm it works for the $\ln 2$ term as well. So, $x = 1$ is a stationary point.
Is $x = 1$ a global maximum? Yes. We can see this by showing the derivative is negative for $x > 1$ and positive for $x < 1$.
First, recall that the map $y \mapsto y^x$ is strictly convex if $x > 1$, and strictly concave if $x < 1$. In particular, for $x > 0$, we have $\frac{2^x + 4^x}{2} > 3^x$, hence $2 \cdot 3^x - 2^x - 4^x < 0$. For $x < 1$, the inequality is reversed. Similarly, when $x > 0$, we have $2 \cdot 2^x - 1 - 3^x < 0$, but when $x < 0$, the inequality is reversed. We also have:$$3 \cdot 4^x - 2 \cdot 6^x = 3 \cdot 4^x\left(1 - \left(\frac{3}{2}\right)^{x-1}\right)$$which is also negative when $x > 1$, and positive when $x < 1$. Thus, putting it all together, we see $f'(x) < 0$ for $x > 1$ and $f'(x) > 0$ for $x < 1$. Thus, $f$ achieves a global maximum at $x = 1$. The value of this maximum is $f(1) = 1$.
