${x^{{x^2} - 3}} + {(x - 3)^{{x^2}}},$ for $x > 3.$

Let $y = {x^{{x^2} - 3}} + {(x - 3)^{{x^2}}} = u + v,$
where $u = {x^{{x^2} - 3}},$ and $v = {(v - 3)^{{x^2}}}$

Let $u = {x^{{x^2} - 3}}$

By taking log on both sides , we get
$\log u = ({x^2} - 3)\log x$ ...(i)

Differentiating (i) w.r.t. x on both sides, we get

$\cfrac{1}{u}\cfrac{{du}}{{dx}} = \cfrac{{({x^2} - 3)}}{x} + \log x(2x)$

therefore, $\cfrac{{du}}{{dx}} = {x^{{x^2} - 3}}\left[ {\cfrac{{{x^2} - 3}}{x} + 2x\log x} \right]$ …(ii)

and $v = {(x - 3)^{{x^2}}}$

By taking log on both sides , we get
$\log v = {x^2}\log (x - 3)$ ….(iii)

Differentiating (iii) w.r.t.x, we get

$\cfrac{1}{v}\cfrac{{dv}}{{dx}} = \cfrac{{{x^2}}}{{x - 3}} + \log (x - 3) \cdot (2x)$

therefore, $\cfrac{{dv}}{{dx}} = {(x - 3)^{{x^2}}}\left[ {\cfrac{{{x^2}}}{{x - 3}} + 2x\log (x - 3)} \right]$ ….(iv)

So, $\cfrac{{dy}}{{dx}} = \cfrac{{du}}{{dx}} + \cfrac{{dv}}{{dx}}$

$= {x^{{x^2} - 3}}\left[ {\cfrac{{{x^2} - 3}}{x} + 2x\log x} \right] + {(x - 3)^{{x^2}}}\left[ {\cfrac{{{x^2}}}{{x - 3}} + 2x\log (x - 3)} \right]$

[from (ii) \& (iv)]