If $$\log_{(x^{2})}y + \log_{(y^{2})} x= 1$$ and $$y = x^{2} - 30$$, then the value of $$x^{2} + y^{2}$$ is _________

Given, $$\log_{(x^{2})}y + \log_{(y^{2})} x= 1$$

or, $$\dfrac{\log\ y}{\log\ x^2}+\dfrac{\log\ x}{\log\ y^2}=1$$
or, $$\dfrac{\log\ y}{2\log\ x}+\dfrac{\log\ x}{2\log\ y}=1$$

or, $$\dfrac{\log\ y}{\log\ x}+\dfrac{\log\ x}{\log\ y}=2$$

or, $$\left(\log\ y\right)^2+\left(\log\ x\right)^2=2\left(\log y\right)\left(\log x\right)$$

or, $$\left(\log\ y\right)^2+\left(\log\ x\right)^2-2\left(\log y\right)\left(\log x\right)=0$$

or, $$\left(\log\ y-\ \log\ x\right)^2=0$$

or, $$\log\ y=\log\ x$$

or, $$x=y$$

Now, $$y=x^2-30$$

or, $$x=x^2-30$$

or, $$x^2-x-30=0$$

or, $$x^2-6x+5x-30=0$$

or, $$\left(x-6\right)\left(x+5\right)=0$$

or, $$x=6$$ or $$x=-5$$

But for $$\log x$$ to be defined, $$x$$ has to be positive

So, $$x=6$$

So, $$x^2+y^2=6^2+6^2=36+36=72$$