If you’re having trouble computing eigenvalues, try looking for eigenvectors instead. You should be able to tell at a glance that $(0,0,1)^T$ is an eigenvector of $H$ since that vector gets mapped to itself—the columns of $H$ are the images of the basis vectors. The corresponding eigenvalue is, of course, $1$.
Now focus on the upper-right $2\times2$ submatrix. Both rows have the same elements, so the row sums are equal. Summing the first two rows of $H$ is equivalent to right-multiplying by $(1,1,0)^T$, so there’s another eigenvector that’s linearly independent of the first one, with eigenvalue $\frac32-\frac12=1$.
You can always get the last eigenvalue “for free” since the sum of the eigenvalues, taking multiplicity into account, is equal to the trace. So, the remaining eigenvalue is $4-1-1=2$. To find a corresponding eigenvector, recall that a real symmetric matrix can be orthogonally diagonalized, so any eigenvector of $2$ has to be orthogonal to both of the eigenvectors of $1$ that you’ve already found. You’re working in $\mathbb R^3$, so such a vector can be found via a cross product: $(1,1,0)^T\times(0,0,1)^T=(1,-1,0)^T$ (which could also have been found by inspection).