Answer: \(y=3x\); \((x,y)=(\sqrt2,3\sqrt2)\) or \((-\sqrt2,-3\sqrt2)\).

Rewrite the exponential terms in a common form, or introduce a substitution, so that the equation becomes algebraic.

Write all terms as powers of \(2\):

\(256^{x+y}\times16^{-2x}=8^{-x+3y}\)

\(2^{8(x+y)}\times2^{4(-2x)}=2^{3(-x+3y)}\).

So

\(2^{8x+8y-8x}=2^{-3x+9y}\).

Hence \(8y=-3x+9y\), giving \(y=3x\).

Substitute \(y=3x\) into \(x^2+3y^2=56\):

\(x^2+3(9x^2)=56\).

Thus \(28x^2=56\), so \(x^2=2\), and \(x=\pm\sqrt2\).

Since \(y=3x\), the exact solutions are

\((x,y)=(\sqrt2,3\sqrt2)\) or \((-\sqrt2,-3\sqrt2)\).

This gives the required answer: \(y=3x\); \((x,y)=(\sqrt2,3\sqrt2)\) or \((-\sqrt2,-3\sqrt2)\).