Let $f:\mathbb{R}^2 \to{\mathbb{R}}$ be s.t $f_x=\frac{x}{\sqrt{x^2-y^2}}$ and $f_y=\frac{y}{\sqrt{x^2-y^2}}$ , $x^2 \ne y^2$
consider the following statements
i) $\lim_{(x,y)\to (2,-1)} f(x,y)$ exists.
ii) f(x,y) is continuous at (2,-1)
then which of the statements is/are correct?
I know that existence of partial derivatives at (2,-1) do not give guarantee of continuity at (2,-1)
Then how I can conclude answer here.
From $f_x=\frac{x}{\sqrt{x^2-y^2}}$ we get $f(x,y)=\sqrt{x^2-y^2}+g(y)$ with a differentiable function $g$ (independent of $x$).
Then: $f_y=\frac{y}{\sqrt{x^2-y^2}}=\frac{-y}{\sqrt{x^2-y^2}}+g'(y)$.
Hence $g'(y)=\frac{2y}{\sqrt{x^2-y^2}}$, which is not independent of $x$, a contradiction.
Conclusion: a function $f$ with the stated properties does not exist.
If $f$ would exist, both were correct since the existence of continuos partial derivatives at a point guarantees differentiability at the point which implies continuity.
Notably for the "Differentiability theorem" if all the partial derivatives exist and are continuous in a neighborhood of the point then (i.e. sufficient condition) the function is differentiable at that point.
Thus it would be
$$\lim_{(x,y)\to (2,-1)} f(x,y)=f(2,-1)$$
Anyway as shown by Fred such f doesn't exists.
If $f_y=\frac{y}{\sqrt{x^2-y^2}}$ is a typo and if it should read $f_y=-\frac{y}{\sqrt{x^2-y^2}}$, then the function $g$ in my first answer is constant.
Hence: $f(x,y)=\sqrt{x^2-y^2}+c$ and $\lim_{(x,y)\to (2,-1)} f(x,y)$ exists and $f$ is continuous at $(2,-1)$.