The given function is
$$ f(x, y) = \frac{1}{x^2+y^2-1} $$
From my understanding, a rational function such as this one would not have any points in which $ \nabla f(x, y) = \mathbf{0} $, so I would conclude that no such points exist. Would this be true or am I not looking at this in a different way?
You need to compute the gradient of $f(x,y)$:
$$ \frac{\partial f}{\partial x} = -\frac{2x}{(x^2 + y^2 - 1)^2}, \quad \frac{\partial f}{\partial y} = -\frac{2y}{(x^2 + y^2 - 1)^2} $$
Hence $\nabla f(x,y) = 0 \iff (x,y) = (0,0)$. Therefore $(0,0)$ is the only stationary point of $f$.
by the quotient rule we get $$f_x=-\frac{2x}{(x^2+y^2-1)^2}$$ $$f_y=-\frac{2y}{(x^2+y^2-1)^2}$$ solve the systemj $$f_x=0,f_y=0$$