Absolute maximum of $f(x) = \sqrt{x^4-3x^2-6x+13}-\sqrt{x^4-x^2+1}$ My maths tutor solved  this by using $f(x) = \sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+(x-0)^2}$ and treating $A(2,3)$ and $B(1,0)$ as fixed points and $P(x^2,x)$ as a moving point and using the difference between the sides PA and PB of the triangle PAB and relating that to the side AB to find the answer. 
i.e. $PA-PB \leq AB = \sqrt{(2-1)^2-(3-0)^2} = \sqrt{10}$
What are other methods to solve this and is there a specific way to turn any such equation into the form of 2 squares as demonstrated above or is that just a special instance wherein that is possible.

No, it's not so hard.
We need to prove that
$$\sqrt{x^4-3x^2-6x+13}-\sqrt{x^4-x^2+1}\leq\sqrt{10}$$ or
$$\sqrt{x^4-3x^2-6x+13}\leq\sqrt{10}+\sqrt{x^4-x^2+1}$$ or
$$1-3x-x^2\leq\sqrt{10(x^4-x^2+1)},$$
which is obvious for $x^2+3x-1\geq0$, but for $x^2+3x-1\leq0$ we need to prove that
$$(x^2+3x-1)^2\leq10(x^4-x^2+1)$$ or
$$9x^4-6x^3-17x^2+6x+9\geq0$$ or
$$(3x^2-x-3)^2\geq0.$$
Done!
