\(\begin{cases}y^{2}-x=-1\\ x=y+3\end{cases}\)

To solve the system of equations:

\[ \begin{cases} y^{2} - x = -1 \\ x = y + 3 \end{cases} \]

we can substitute the second equation into the first equation.

1. From the second equation, we have \(x = y + 3\).
2. Substitute \(x\) in the first equation:

\[ y^{2} - (y + 3) = -1 \]

3. Simplify the equation:

\[ y^{2} - y - 3 + 1 = 0 \] \[ y^{2} - y - 2 = 0 \]

4. Now, we can factor this quadratic equation:

\[ (y - 2)(y + 1) = 0 \]

5. Setting each factor to zero gives us the solutions for \(y\):

\[ y - 2 = 0 \quad \Rightarrow \quad y = 2 \] \[ y + 1 = 0 \quad \Rightarrow \quad y = -1 \]

6. Now, we will substitute these \(y\) values back into the equation \(x = y + 3\) to find the corresponding \(x\) values.

For \(y = 2\):

\[ x = 2 + 3 = 5 \]

For \(y = -1\):

\[ x = -1 + 3 = 2 \]

7. Therefore, the solutions to the system of equations are:

\[ (5, 2) \quad \text{and} \quad (2, -1) \]

The final result is: \[ \boxed{(5, 2) \text{ and } (2, -1)} \]