b) \( \frac{y^{2}}{y^{2}-4}=\frac{5 y-6}{y^{2}-4} \)

To solve the equation \( \frac{y^{2}}{y^{2}-4}=\frac{5y-6}{y^{2}-4} \), we can start by observing that both sides have a common denominator \( y^{2}-4 \) (as long as \( y^{2} \neq 4 \)). We can multiply both sides by this denominator to eliminate it: \[ y^{2} = 5y - 6. \] Next, we can rearrange the equation into standard quadratic form: \[ y^{2} - 5y + 6 = 0. \] Now we can factor the quadratic: \[ (y - 2)(y - 3) = 0. \] Setting each factor equal to zero gives us the solutions: \[ y - 2 = 0 \quad \Rightarrow \quad y = 2, \] \[ y - 3 = 0 \quad \Rightarrow \quad y = 3. \] However, we must ensure these values do not make the original denominator zero. The denominator \( y^{2}-4 \) becomes zero when \( y = 2 \) or \( y = -2 \). Since \( y = 2 \) makes the denominator zero, we must exclude it from our solutions. Thus, the only solution to the original equation is: \[ y = 3. \]