Double angle identity for cosine:
[tex]\cos\left(2\sin^{-1}\dfrac35\right)=\cos^2\left(\sin^{-1}\dfrac35\right)-\sin^2\left(\sin^{-1}\dfrac35\right)[/tex]
By the Pythagorean theorem, if [tex]\sin x=\dfrac35[/tex], it follows that [tex]\cos x=\dfrac45[/tex]:
[tex]\cos^2x+\sin^2x=1\implies \cos x=\pm\sqrt{1-\sin^2x}[/tex]
where we take the positive root because [tex]-\dfrac\pi2\le\sin^{-1}x\le\dfrac\pi2[/tex], and [tex]\cos x[/tex] is non-negative throughout this domain.
So we have
[tex]\cos^2\left(\sin^{-1}\dfrac35\right)-\sin^2\left(\sin^{-1}\dfrac35\right)=\left(\dfrac45\right)^2-\left(\dfrac35\right)^2=\dfrac7{25}[/tex]
