Differentiate the function with respect to $x$: $x^{x^{2}-3}+(x-3)^{x^{2}}$ for $x > 3$.

Let $y = x^{x^{2}-3} + (x-3)^{x^{2}}$.
Let $u = x^{x^{2}-3}$ and $v = (x-3)^{x^{2}}$.
Then $y = u + v$,so $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ ... $(1)$.
For $u = x^{x^{2}-3}$,taking log on both sides: $\log u = (x^{2}-3) \log x$.
Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = (x^{2}-3) \cdot \frac{1}{x} + \log x \cdot (2x)$.
Thus,$\frac{du}{dx} = x^{x^{2}-3} \left( \frac{x^{2}-3}{x} + 2x \log x \right)$.
For $v = (x-3)^{x^{2}}$,taking log on both sides: $\log v = x^{2} \log (x-3)$.
Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = x^{2} \cdot \frac{1}{x-3} + \log (x-3) \cdot (2x)$.
Thus,$\frac{dv}{dx} = (x-3)^{x^{2}} \left( \frac{x^{2}}{x-3} + 2x \log (x-3) \right)$.
Substituting these into $(1)$,we get $\frac{dy}{dx} = x^{x^{2}-3} \left( \frac{x^{2}-3}{x} + 2x \log x \right) + (x-3)^{x^{2}} \left( \frac{x^{2}}{x-3} + 2x \log (x-3) \right)$.