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I want to show that $f(x,y)=(2x^2-y)(y-x^2)$ has positive and negative function values in every neighborhood around $(0,0)$.

Is my reasoning sufficient to prove this claim?

On the one hand, I approached $(0,0)$ using $y=\sqrt{2}x^2$. Then, for all $x\in \mathbb{R}\setminus \{0\}$, we have:$$f(x,\sqrt{2}x^2)=(3\sqrt{2}-4)x^4>0$$ and on the other hand, I approached $(0,0)$ using $x=0$. Then, for all $y\in \mathbb{R}\setminus \{0\}$, we have:$$f(0,y)=-y^2<0$$

My reasoning is that you can get arbitrarily close to $(0,0)$ using both ways to approach $(0,0)$ which guarantee negative and positive function values. Is this sufficient?

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    $\begingroup$ Somewhat off-topic: just because I had trouble following your logic does not imply that your approach is invalid. It could easily be my shortcoming. Anyway, my approach is to have $x$ (and therefore $x^2$) approach $0$, with $y_1$ being selected as $< x^2$, and $y_2$ being selected so that $x^2 < y_2 < 2x^2$. Clearly, in any neighborhood around $(0,0)$, you will be able to select suitable $x, y_1, y_2$. $\endgroup$
    – user2661923
    Commented Aug 23, 2021 at 1:15
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    $\begingroup$ I just checked on symbolic calculator, your argument is OK. [There may be a faster way to get positive values...] $\endgroup$
    – coffeemath
    Commented Aug 23, 2021 at 1:18
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    $\begingroup$ @Ugo But $f(0,-1)=-1,$ and generally $f(0,y)=-y^2$ whether $y$ is positive or negative. OP already said that. $\endgroup$
    – coffeemath
    Commented Aug 23, 2021 at 1:45
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    $\begingroup$ @Ugo Yes, in particular note $f(t,\frac{3}{2}t^2)=\frac{t^4}{4}.$ That looks a bit simpler than OP's example, but can't just multiply $x,y$ by 2 here since $f$ isn't homogeneous, in fact then it would give a negative. $\endgroup$
    – coffeemath
    Commented Aug 23, 2021 at 5:05
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    $\begingroup$ @Ugo Just noticed a slight simplification: $f(2t,5t^2)=3t^4.$ $\endgroup$
    – coffeemath
    Commented Aug 23, 2021 at 5:12

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