Continuity and differentiability of $f(x, y) = \frac{2x^3 + 3y^3}{x^2 + y^2}$

Your part (a) is perfect!

For differentiability a necessary condition is that a directional derivative should exist and depend linearly on the direction: Given $(h,k)$ we calculate $$ Df_{(0,0)} (h,k) = \frac{d}{dt}_{|t=0} f(th,tk) = \frac{d}{dt}_{|t=0} t f(h,k) = f(h,k) = \frac{2h^3+3k^3}{h^2+k^2}$$ The directional derivative exists in any direction but this derivative is not linear as a function of $h$, $k$ (check why!). So the function $f$ is not differentiable at zero.

Note that a sufficient (though not necessary) condition is that partial derivatives exist and are continuous (not the case here).

To complete the story: Suppose (hypothetically) that you had found a directional derivative which is linear in $(h,k)$, thus $Df_{(0,0)} (h,k) =ah+bk$ for some $a$ and $b$. Then there are still some hurdles, as you would have to show that the error term is small (called little 'o') compared to the length of $(h,k)$. More precisely, that $$ \lim_{h^2+k^2\rightarrow 0} \frac{f(h,k)-ah-bk}{\sqrt{h^2+k^2}} = 0$$ Luckily enough you have not arrived at this...

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