Solution of $\phi \left( x \right) = f{\left( x \right)^{g\left( x \right)}}$

$$\left ( x^2-7x+11 \right )^{\left ( x^2-13x+42 \right )}=1\Leftrightarrow \left ( x^2-7x+11 \right )^{\left ( x^2-13x+42 \right )} =\left ( x^2-7x+11 \right )^0$$ $$\left ( x^2-7x+11-1 \right )\left ( x^2-13x+42 \right )=0\Leftrightarrow \left ( x^2-7x+10 \right )\left ( x^2-13x+42 \right )=0$$ $$x^2-7x+10=0\Rightarrow x=\left \{ 2,5 \right \}$$ $$x^2-13x+42=0\Rightarrow x=\left \{ 6,7 \right \}$$ You need to check the roots: $$x^2-7x+11>0$$ All four roots fit!

Eric is correct, but you need to exclude the case where $|f(x)|=0$, because $0^{0}$ is indeterminate form. $f(x)$ can be negative, as long as $g(x)$ makes $\phi(x)$ defined and real, in our case as long as $g(x)$ (even).