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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.

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    i would compute the first derivative – 2017-01-08
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    I've tried it results in an equation too difficult to solve. – 2017-01-08
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    ok then we can use a numerical method – 2017-01-08
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    $(x^2-a)^2+(x-b)^2=x^4+(1-2a)x^2-2bx+(a^2+b^2)$, so the technique your tutor used seems rather limited in applicability. – 2017-01-08
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    I am certain that the person who wrote this problem meant it to be solved by your tutor's trick method only. Solving it via first derivative test will yield an equation of degree 10 and that is, in general, not solvable. – 2017-01-08
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    @dezdichado. The problem reduces to an equation of degree 7 which has two "simple" roots $x=\frac{1\pm\sqrt{37}}{6}$ !! But the problem is still there. Cheers. – 2017-01-08
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    One can maybe generalize.. Any excursion function associating to an arbitrary self cancelling curve $y=E(x)$ generated distances inside a function $f(x)$ $$ y=f(x)= \sqrt{(E(x)-x_a)^2 + (x-y_a)^2)} - \sqrt{(E(x)-x_b)^2 + (x-y_b)^2)} $$ which we purposely set up knowing well *a priori* minimum Euclidean distance between the fixed points... I think that needs no further verification. However routine differentiation procedure confirms the result. – 2017-01-08
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    contd.. What I mean is that if we use $ P( 2- x, x) $ instead of $P(x^2,x)$ the problem set up is different, but the result is same. Similarly for $f(x) = \sqrt{(e^x-2)^2+(x-3)^2}-\sqrt{(e^x-1)^2+(x-0)^2}$ and treating $A(2,3)$ and $B(1,0)$ as fixed points for excursor point $P(e^x,x).$ – 2017-01-08

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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!