Let's redraw the schematic:
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simulate this circuit– Schematic created using CircuitLab
Ignore \$C\$ on the left side. Given enough time it's an open circuit. Here KCL says \$\frac{v_i}{R}=\beta\,v_o\$ or \$v_i=\beta\,R\,v_o\$.
Ignore \$C\$ on the right side. Given enough time it's also an open circuit. Here KCL says \$\frac{v_o}{R}=-\beta\,v_o\$ or \$\beta=-\frac1{R}\$.
Since the left side node voltage depends directly and entirely upon the right side node voltage, there's no point looking at the left side. It's a red herring.
Returning to the right side and thinking about when \$\beta=-\frac1{R}\$. Suppose \$v_o\$ moves more positive. Then \$\frac{v_o}{R}\$ increases and this is exactly met by \$I_y\$. But \$C\$ on the right also requires a current pointing in the same direction as \$\frac{v_o}{R}\$ and there's no source for that current. KCL is violated. So as a result of adding \$C\$ we know that \$\beta\ne -\frac1{R}\$.
The only way that there can be at least some current to supply the right side \$C\$ if the node voltage changes is to ensure that \$\frac{v_o}{R}\gt-\beta\,v_o\$. Or, \$\beta\gt -\frac1{R}\$.
But this also guarantees that on the right side \$v_o\to 0\:\text{V}\$, given time. And that will eventually also then mean \$v_i=0\:\text{V}\$, too.
Of course, we could have simply recognized that \$v_o= 0\:\text{V}\$ and \$v_i=0\:\text{V}\$ is and always was a solution.
And that's stable.
(Note to self -- it's possible to add an answer to a question an hour later than when the question was closed. Might be a profound truth buried in that fact.)
