Finding continuity and differentiability of a multivariate function Determine whether the following functions are differentiable, continuous, and whether its partial derivatives exists at point $(0,0)$:
(a) $$f(x, y) = \sin x \sin(x + y) \sin(x − y)$$
(b)$$f(x,y)=\sqrt{|xy|}$$
(c)$$f(x, y) = 1 − \sin\sqrt{x^2 + y^2}$$
(d) $$f(x,y) = \begin{cases} \dfrac{xy}{x^2+y^2} & \text{if $x^2+y^2>0 $} \\ 0 & \text{if $x=y=0$} \end{cases}$$
(e) $$f(x,y) = \begin{cases} 1 & \text{if $x y \ne 0$} \\ 0
 & \text{if $xy=0$} \end{cases}$$
(f)$$f(x,y) = \begin{cases} \dfrac{x^2-y^2}{x^2+y^2} & \text{if $x^2+y^2>0$} \\ 0 & \text{if $x=y=0$} \end{cases}$$
My try:
For (a), using the definition of the derivative for a multivariate function, the limit tends to $0$, hence it's differentiable and its partial derivative exists and it's continuous.
For (b) I mentioned that is not differentiable as using the definition of the derivative for a multivariate function, the limit does not tend to $0$. While it's continuous, as the limit of the function $f(x,y)$ tends to $0$. To determine whether its partial derivative exists, this part is tricky because of the modulus sign in the function hence I'm unsure whether a modulus is differentiable for this case.
For part (c) it should be continuous but not differentiable at $(0,0)$ because its partial derivative does not exists at $(0,0)$. As to why its partial derivatives does not exists, lets say to find $$f_x$$, we let $y=0$, the expression $f(x,y)$ becomes $1-\sin(|x|)$ which is not differentable at $(0,0)$, hence partial derivatives cannot exists at $(0,0)$.
For (d), this question is also tricky, as although initially I though its partial derivatives exists, now I think likewise. Because if I want to differentiate the function with respect to $x$ for example, I would sub in the value of $y=0$ making the numerator a zero and hence assume the derivative is $0$. However, on closer look, there is still the denominator of $x^2$ and if $x^2=0$ the denominator becomes $0$ and since $0/0$ is undefined, the partial derivatives do not exist. As for continuity, it is not continuous and hence not differentiable.
For (e) Not differentiable, Discontinuous, Partial derivatives defined
(because "not continuous" will mean it's not differentiable, but I'm unsure of the Partial derivatives portion though because it appears the the partial derivatives are $0$ but I also have the feeling that the partial derivatives do not exist.)
For (f) Not differentiable, Continuous, Partial derivatives defined. This question appears to be similar to question (d)
I have already attempted these questions many times, but I keep answering these questions incorrectly. I know that I must be missing out on some parts especially when these are tricky questions which are not as simple it might seem. Could anyone help me please? Thanks!
 A: (a) it is a compostion of  diferentiable functions, then it is differentiable, and contious and the partial derivative exist in $(0,0)$.
(b) It is continuous,and we have that the partial derivative are 
$f_x(0,0)=\displaystyle\lim_{t\to 0}\frac{f(t,0)-f(0,0)}{t}=
\lim_{t\to 0}\frac{\sqrt{|t0|}-\sqrt{|0\times0|}}{t}=0$, and 
$f_y(0,0)=\displaystyle\lim_{t\to 0}\frac{f(0,t)-f(0,0)}{t}=
\lim_{t\to 0}\frac{\sqrt{|t0|}-\sqrt{|0\times0|}}{t}=0$. however it is not differentiable since $\lim_{t\to0^+}\frac{f(t,t)-f(0,0)}{t}=1$  and $\lim_{t\to0^-}\frac{f(t,t)-f(0,0)}{t}=-1$,then it is not diferentiable.
(c) It is continuous  because it is composition of continuous functions,but
$f_x(0,0)=\displaystyle\lim_{t\to 0^+}\frac{f(t,0)-f(0,0)}{t}=
\lim_{t\to 0^+}\frac{1 − \sin\sqrt{t^2 + 0^2}- (1 − \sin\sqrt{0})}{t}=-1$ but
$f_x(0,0)=\displaystyle\lim_{t\to 0^-}\frac{f(t,0)-f(0,0)}{t}=
\lim_{t\to 0^-}\frac{1 − \sin\sqrt{t^2 + 0^2}- (1 − \sin\sqrt{0})}{t}=1$
and the partial $f_x$ is not defined in $(0,0)$, analogously for $f_y(0,0)$, both does not exist.
Then it is not differentiable, because a differentiable function the elimites above should exist.
(d) it is not continuous, because $t>0$ then $(t,t)\to 0$ then $f(t,t)=1/2\neq 0=f(0,0)$.  And it is not differentiable since it is not continuous. However 
$f_x(0,0)=\displaystyle\lim_{t\to 0}\frac{f(t,0)-f(0,0)}{t}=
\lim_{t\to 0^+}\frac{\dfrac{t0}{t^2+0^2}-0}{t}=0$ and
$f_y(0,0)=\displaystyle\lim_{t\to 0}\frac{f(0,t)-f(0,0)}{t}=
\lim_{t\to 0^+}\frac{\dfrac{t0}{t^2+0^2}-0}{t}=0$.
(e) It is clearly not continuous, hence not differentiable at $(0,0)$, but
$f_x=\displaystyle\lim_{t\to0}\frac{f(x+t,y)-f(x,t)}{t}=0$ and
$f_y=\displaystyle\lim_{t\to0}\frac{f(x,y+t)-f(x,t)}{t}=0$, are defined in $(0,0)$
(f)It is not continuous since $\lim_{t\to 0}f(2t,t)=\lim_{t\to0}\dfrac{4t^2-t^2}{4t^2+t^2}=\frac{3}{5}\neq f(0,0)$, hence it is not differentiable in $(0,0)$.
$f_x(0,0)=\displaystyle\lim_{t\to0^+}\frac{f(x+t,y)-f(x,t)}{t}
=\lim_{t\to 0^+}\frac{\dfrac{t^2-0^2}{t^2+0^2}-0}{t}=+\infty
$ analogously for $f_y(0,0)$, both are note defined in $(0,0)$.
