Continuity of a 2 variable function - Munkres exercise Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ be defined as $f(0,0)=0$
 and $f(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2}$ for $(x,y)\neq (0,0)$
Then question asks to prove that $f$ is differentiable.
Hint that is given is :

Show that $D_1f$ equals product of $y$ and a bounded function and $D_2f$  equals product of $x$ and a bounded function.

I calculated $D_1f=y\frac{x^4+4x^2y^2-y^4}{(x^2+y^2)^2}$ and clearly
$\left|\frac{x^4+4x^2y^2-y^4}{(x^2+y^2)^2}\right|\leq 3$ So, we have $D_1f$ as product of $y$ and a bounded function.. Similarly $D_2f$ is product of $x$ with a bounded function..
I know that if $D_1f$ and $D_2f$ are bounded then $f$ is continuous..
But here it is product of a bounded function with $y$..
I do not know how to proceed..
Please help me...
P.S : I am supposed to prove that it is differentiable.
I think i should use condition that if partial derivatives are continuous then $f$ is differentiable...
 A: Differentiability (using continuity of partial derivatives): Check that both partial derivatives of $f$ at $(0,0)$ are zero. So you want to show $D_1f(x,y) \to 0$ as $(x,y) \to (0,0).$ But you've shown $D_1f(x,y) = y\cdot g(x,y),$ where $g$ is bounded. That implies what you want, right? Same for $D_2f(x,y).$
We can also show $Df(0,0)$ exists more directly. We know both partial derivatives equal $0$ at $(0,0).$ So if $Df(0,0)$ exists, it is the zero linear transformation. So we want to show
$$f(x,y) = f(0,0) + 0\cdot x + 0\cdot y + o(\sqrt {x^2+y^2}) = o(\sqrt {x^2+y^2}).$$
The estimate $|f(x,y)| \le |xy|$ (from below) gives this.

Previous answer on continuity:
Just use
$$|f(x,y)| \le |xy|\frac{x^2 + y^2}{x^2+y^2} = |xy|.$$
We know $xy\to 0$ as $(x,y) \to (0,0)$ and that gives continuity at $(0,0).$
A: For $p=(x,y)$ and $r>0$ let $B(p,r)=\{(x',y'):(x'-x)^2+(y'-y)^2<r^2\}.$ Suppose $g_1:S_1\to R$ and $g_2:S_2\to R$ are differentiable, where $S_1\subset R^2\supset S_2,$ and $S_1\cap S_2\supset B(p,r)$. Then $\bullet$(1): $g_1,g_2, g_1+g_2,$ and $g_1 g_2$ are differentiable on $B(p,r).$  $\bullet$(2): If  $g_1 \neq 0$ on $B(p,r)$ then $g_2/g_1$ is differentiable on $B(p,r).$..... In your Q, with $S_1=S_2=R^2\backslash \{(0,0)\}$ and $p=(x,y)\ne (0,0),$ apply (1) and (2) repeatedly with $r=\sqrt {x^2+y^2}.$
