Isn't this a contradiction to the zero partial derivative necessary condition for an extremum?

Question

I learned that for a point to be an extremum of a function, the necessary condition is that the partial derivatives of the function with respect to its variables must be zero. However, suppose I have a function $$f(x,y(x)) = x - \sqrt{y}$$ and $y$ implicitly depends on $x$ such that $y(x) = x$.

The minimum of this function is achieved at $x = 1/4$.

However, $\frac{\partial f}{\partial x} = 1$ for all $x$.

Isn't this a contradiction?

Answer 1

In the statement

for a point to be an extremum of a function, the necessary condition is that the partial derivatives of the function with respect to its variables must be zero

the variables must be independent.

Answer 2

You are right. The partial derivative with respect to $x$ is indeed $1$. Sorry, I made it wrong before.

However, the main difference here is that we have a restriction on $y$. Globally $x=1/4$ is not an extremum or there is even no global extremum for $f$, because $f_x=1$ is a constant.

$x=1/4$ is indeed the extremum on the intersection of curve plane $z=x-\sqrt{y}$ and plane $y=x$. The intersection, however, is a simple curve. It would be absurd to consider any partial derivative on a curve. In other words, we cannot even apply this principle, nor to prove it wrong.