$\bf{Definition:}$ We say that $f$ not depends of $i$-th variable if $x, x + te_{i} \in U \Longrightarrow f(x) = f(x + te_{i})$.
Let $X = \lbrace (x,0)\;|\;x \geq0\rbrace$ and $U = \mathbb{R}^{2} - X$. Define $f:U \longrightarrow \mathbb{R}$ by $f(x,y) = x^{2}$ if $x > 0, y>0$ and $f(x,y) = 0$ if $x \leq 0$. Show that $\displaystyle \frac{\partial f}{\partial y} = 0$ for all $u \in U$ but $f$ depends of $y$.
$\it{Partial\; solution}$ (Attemp). If $x_{1}>0, y_{1}>0$
$$\displaystyle \frac{\partial f}{\partial y}(x_{1},y_{1}) = \lim_{t\to 0}\frac{f(x_{1}, y_{1} + ty) - f(x_{1},y_{1})}{t} = \lim_{t\to 0}\frac{x_{1}^{2} - x_{1}^{2}}{t} = 0$$
and if $x_{1} \leq 0$ $$\displaystyle \frac{\partial f}{\partial y} = \lim_{t\to 0}\frac{f(x_{1}, y_{1} + ty) - f(x_{1},y_{1})}{t} = \lim_{t\to 0}\frac{0 - 0}{t} = 0.$$
But, and $y<0$? The function is defined? Moreover, I don't see how $f$ depends of $y$, because only $x$ determines de value of the function (intuitively). Can anybody help me?
The domain $U\ne\Bbb R^2$. First, $f$ isn't defined when $x\ge 0$ and $y = 0$. In particular, $f(0,0)$ isn't defined... but after is defined when you say $f(x,y) = 0$ if $x\ge 0$.
Another problem: You don't define $f$ for all the points of $U$. Namely, for $x^2 > 0$, $y < 0$ .
Check again.