Given \( x*y=x^2+y^3 \)

\(\therefore (x*1)*1=x*(1*1)\)

\((x^2+1)*1=x*(2)\)

\((x^2+1)^2+1=x^2+8\)

\(x^4+2x^2+1+1=x^2+8\)

\(x^4+x^2-6=0\)

\((x^2+3)(x^2-2)=0\)

\(x^2=2\)

Now,

\(2\sin^{-1}\left(\frac{x^4+x^2-2}{x^4+x^2+2}\right)\)

\(=2\sin^{-1}\left(\frac{4+2-2}{4+2+2}\right)\)

\(=2\sin^{-1}\left(\frac{1}{2}\right)\)

\(=2\cdot\frac{\pi}{6}\)

\(=\frac{\pi}{3}\)