This is not an answer (as Robert has already provided one), but gives a hint at how the $w$ affects the eigenvalues (and doesn't fit in the comments) in some particular cases.
If $U^T YU=\Lambda$, where $\Lambda$ is diagonal with elements $\lambda_1,...,\lambda_n$, then $\begin{bmatrix} U^T & 0 \\ 0 & I \end{bmatrix} \begin{bmatrix} Y & -w \\ -w^T & b \end{bmatrix} \begin{bmatrix} U & 0 \\ 0 & I \end{bmatrix} = \begin{bmatrix} \Lambda & -Uw \\ -w^T U & b \end{bmatrix}$. Let $\tilde{w} = Uw$ for convenience.
Suppose $x \notin \{\lambda_1,...,\lambda_n\}$, then we have \begin{eqnarray} \det(\begin{bmatrix} xI-\Lambda & \tilde{w} \\ \tilde{w}^T & x-b \end{bmatrix}) &=& \det (xI-\Lambda)(x-b - \tilde{w}^T (xI-\Lambda)^{-1} \tilde{w}) \\ &=& \det (xI-\Lambda)(x-b - \sum_i \frac{\tilde{w}_i^2}{x-\lambda_i}) \\ &=& (x-\lambda_1)\cdots(x-\lambda_n)(x-b)-\sum_i \tilde{w}_i^2 \prod_{j \neq i}(x -\lambda_j) \end{eqnarray} Hence $\det(xI-Y) = (x-\lambda_1)\cdots(x-\lambda_n)(x-b)-\sum_i \tilde{w}_i^2 \prod_{j \neq i}(x -\lambda_j)$.