Is there a relationship between slope of the curve $ f(x, y) = 0 $ and the partial derivative $ \frac{\partial f}{\partial x} $? Let $ f(x, y) = 0 $ define a curve. Is there any relationship between the slope of this curve and the partial derivative $ \frac{\partial f}{\partial x} $?
For example, if $ f(x, y) = (x - 1)^2 + (y - 1)^2 - 1 $, then $ f(x, y) = 0 $ defines a circle of radius $ 1 $ centered at $ (0, 0) $.
Now $ \frac{\partial f(x, y)}{\partial x} = 2(x - 1) $.
Thus $ \frac{\partial f}{\partial x} = 0 $ at $ x = 1 $ and indeed we see that the circle $ f(x, y) = 0 $ also has a slope of $ 0 $ at $ x = 1 $.
Is it always true that $ \frac{\partial f}{\partial x} = 0 $ at $ x = a $ implies that the curve $ f(x, y) = 0 $ has a horizontal slope (slope = $ 0 $) at $ x = a $?
If yes, how can we prove it?
If not, is there a counterexample?

$\def\rbf{\mathbf{R}}$The equation $f(x,y)=0$ does not always define a (smooth) curve. But let us assume, for simplicity, that it does and $f:\rbf^2\to\rbf$ is smooth and the gradient $\nabla f\ne 0$ everywhere.
It does not make sense to say "the slope of this curve". Though it makes sense to say the slope field on this curve or the slope of the tangent line at a point of this curve.
Note $f=0$ is a level curve for the function $f$. Call this curve $\gamma$.
It is known that

At each point on this level curve, the gradient vector is perpendicular to the level curve.

In other words, if $(a,b)$ is a point on the curve, i.e., $f(a,b)=0$, and $f_x(a,b)=A$ and $f_y(a,b)=B$, then an equation to the tangent line of $\gamma$ at $(a,b)$ is
$$
A(x-a)+B(y-b)=0\;\tag{1}
$$
So if $f_x(a,b)=0$, equation (1) becomes
$$
B(y-b)=0\;.
$$
which is a horizontal line ($B\ne 0$ by our assumption).

[Added.] As one of the comments pointed out, in your example $f(x,y)=(x-1)^2+(x-1)^2-1$, the equation $f=0$ is corresponding to the circle centered at $(1,1)$, not $(0,0)$.
Moreover, it does not make sense to say the circle $f=0$ has a slope of $0$ at $x=1$. What you have instead, is that at the point $(x,y)=(1,0)$ and also the point $(x,y)=(1,2)$, the tangent lines to the circle both have the slope $0$.
