The given equation can be rewritten as \(9^{y} - 4 \times 3^{y} + 3 = 0\). Let's substitute \(x = 3^{y}\) to get a quadratic equation in terms of x: \(x^2 - 4x + 3 = 0\).

This quadratic equation can be factored as \((x - 1)(x - 3) = 0\), which has solutions \(x = 1\) and \(x = 3\). 

Substituting back \(x = 3^{y}\), we get two equations: \(3^{y} = 1\) and \(3^{y} = 3\).

The first equation has solution \(y = 0\) and the second equation has solution \(y = 1\). 

Therefore, the values of y which satisfy the given equation are 0 and 1.