<s>[INST] Continuity and differentiability of $f(x, y) = \frac{2x^3 + 3y^3}{x^2 + y^2}$ I'm learning multivariable calculus, specifically multivariable continuity and differentiability, and need help with the following exercise:

Let
$$f(x,y) = \begin{cases} \frac{2x^3 + 3y^3}{x^2 + y^2}, & (x,y) \neq (0, 0) \\ 0, & (x,y) = (0, 0).  \end{cases}$$
$(a)$ Examine if $f$ is continuous at $(0, 0)$.
$(b)$ Examine if $f$ is differentiable at  $(0, 0)$.

Since I'm having difficulties for $(b)$, I'm going to show my work for $(a)$.
$(a)$ We note that
$$0 \leq \left| \frac{2x^3 + 3y^3}{x^2 + y^2}  \right| \leq \frac{2|x|^3}{x^2 + y^2} + \frac{3|y|^3}{x^2 + y^2} \leq \frac{2|x|^3}{x^2} + \frac{3|y|^3}{y^2} = 2|x| + 3|y|$$
and $$\lim_{(x,y) \to (0, 0)} 2|x| + 3|y| = 0.$$
Therefore, by the Squeeze theorem, we conclude that
$$\lim_{(x, y) \to (0, 0)} f(x, y) = f(0, 0)$$
that is
$$\lim_{(x, y) \to (0, 0)} \frac{2x^3 + 3y^3}{x^2 + y^2} = 0$$
so $f$ is continuous at $(0, 0)$.

Is my work correct for $(a)$? How do I examine differentiability of $f$ at $(0, 0)$? More importantly: what is the method to examine differentiability in this case? What do I need to show?
 [/INST] Your proof about continuity  is correct.
In order to check differentiability fistly find partial derivatives so that 
$$ { f }_{ x }^{ \prime  }\left( 0,0 \right) =\lim _{ x\rightarrow 0 }{ \frac { f\left( x,0 \right) -f\left( 0,0 \right)  }{ x }  } =\lim _{ x\rightarrow 0 }{ \frac { \frac { 2x^{ 3 } }{ x^{ 2 } }  }{ x }  } =2\\ { f }_{ y }^{ \prime  }\left( 0,0 \right) =\lim _{ y\rightarrow 0 }{ \frac { f\left( y,0 \right) -f\left( 0,0 \right)  }{ x }  } =\lim _{ x\rightarrow 0 }{ \frac { \frac { 3y^{ 3 } }{ y^{ 2 } }  }{ y }  } =3$$
differentiability should be shown as

$$f\left( x,y \right) -f\left( 0,0 \right) =\frac { 2x^{ 3 }+3y^{ 3 } }{ x^{ 2 }+y^{ 2 } } =2x+3y+\left( \frac { 2x^{ 3 }+3y^{ 3 } }{ x^{ 2 }+y^{ 2 } } -2x-3y \right) =\\={ f }_{ x }^{ \prime  }\left( 0,0 \right) x+{ f }_{ y }^{ \prime  }\left( 0,0 \right) y+\alpha \left( x,y \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } $$ where $$\alpha \left( x,y \right) =\frac { -3{ x }^{ 2 }y-2{ y }^{ 2 }x }{ \left( x^{ 2 }+y^{ 2 } \right) \sqrt { x^{ 2 }+y^{ 2 } }  } $$

We should check whether is this infinity small?
Let $x=\frac { 1 }{ n } ,y=\frac { 1 }{ n } $ so that 

$$\\ \alpha \left( \frac { 1 }{ n } ,\frac { 1 }{ n }  \right) =\frac { -\frac { 5 }{ { n }^{ 3 } }  }{ \frac { 2\sqrt { 2 }  }{ { n }^{ 3 } }  } =-\frac { 5 }{ 2\sqrt { 2 }  } \neq 0,n\rightarrow \infty ,n\in \ \mathbb{N} $$

it shows $\alpha \left( x,y \right) $ is not  infinitly small (when $x\rightarrow 0,y\rightarrow 0$) in other words

$$\alpha \left( x,y \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \neq o\left( \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  \right) $$ 

so it not differentiable
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