= Solution
Differentiability at $x\in\mathbb R^2$ means that there is a <linear map> $Df|_x:\mathbb R^2\to\mathbb R$ such that
$$
f(x+h)=f(x)+Df|_x(h)+o(\|h\|).
$$
The <partial derivatives> are $D_if(x)=Df|_x(e_i)$ when the <derivative> exists; independently, they may be defined by the corresponding one-variable difference quotients.
Suppose the <partial derivatives> exist near $x=(x_1,x_2)$ and are continuous at $x$. Apply the one-dimensional mean value theorem along the two coordinate segments from $x$ to $x+h$. It gives intermediate points $\xi_h,\eta_h\to x$ such that
$$
f(x+h)-f(x)
=D_1f(\xi_h)h_1+D_2f(\eta_h)h_2.
$$
Continuity of the partials makes the remainder after subtracting
$$
D_1f(x)h_1+D_2f(x)h_2
$$
equal to $o(\|h\|)$. Hence $f$ is <differentiable> and this is its <derivative>.
For the given <function>, writing $r=\sqrt{x^2+y^2}$, at $r>0$ we have
$$
\boxed{
D_1f=2x\sin(1/r)-\frac{x}{r}\cos(1/r),
\qquad
D_2f=2y\sin(1/r)-\frac{y}{r}\cos(1/r)}.
$$
At the origin both <partial derivatives> are zero, since $f(h,0)/h$ and $f(0,h)/h$ tend to zero. Neither partial is continuous there: along its corresponding coordinate axis the cosine term oscillates without a <limit>. Nevertheless
$$
|f(x,y)|\leq r^2=o(r),
$$
so $f$ is <differentiable> at the origin with <derivative> zero. This illustrates that <existence of partial derivatives does not imply their continuity>.
The final assertion is false. Define
$$
g(x,y)=
\begin{cases}
(x^2+y^2)\sin\!\left(\dfrac1{x^2+y^2}\right),&(x,y)\ne(0,0),\\
0,&(x,y)=(0,0).
\end{cases}
$$
Again $|g|\leq r^2$, so $g$ is <differentiable> at the origin and it is smooth elsewhere. But along the $x$-axis,
$$
D_1g(x,0)=2x\sin(1/x^2)-\frac2x\cos(1/x^2),
$$
is unbounded near zero, and similarly $D_2g(0,y)$ is unbounded. Thus both <partial derivatives> can be unbounded in every neighbourhood of a point even when the <function> is <differentiable> everywhere.
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