Switch to polar coordinates $x=r\cos\theta$, $y=r\sin\theta$, so $x^2+y^2=r^2$ and for $r\neq 0$, \[f(r,\theta)=r^2\sin\frac{1}{r^2}.\]
Continuity: Since $\left|r^2\sin(1/r^2)\right|\le r^2$ for every $\theta$, the squeeze theorem gives $f\to 0$ as $r\to 0$, independent of direction. As $f(0,0)=0$, f is continuous at the origin.
Linear approximation: Write $f(x,y)$ near the origin as \[f(x,y)=0+0\cdot x+0\cdot y+\varepsilon(r)\cdot r,\] where $\varepsilon(r)=r\sin(1/r^2)$ for $r\neq 0$ and $\varepsilon(0)=0$. This is the affine approximation with both partial derivatives equal to 0 at the origin, plus a remainder $\varepsilon(r)\cdot r$. For f to be differentiable at (0,0), it suffices that $\varepsilon(r)\to 0$ as $r\to 0$. Since $|\varepsilon(r)|=|r\sin(1/r^2)|\le r\to 0$, this holds, so the affine map $L(x,y)=0$ is the derivative of f at (0,0), i.e. f is differentiable there with $df_{(0,0)}=0$.
The partial derivative functions off the origin: For $(x,y)\neq (0,0)$, \[f_x(x,y)=2x\sin\frac{1}{r^2}-\frac{2x}{r^2}\cos\frac{1}{r^2},\] and along $y=0$, $x=r$, this becomes $2r\sin(1/r^2)-\frac{2}{r}\cos(1/r^2)$, which oscillates without bound as $r\to 0$ (choosing $r^2=1/(2n\pi)$ makes $\cos(1/r^2)=1$ while $2/r\to\infty$). So $f_x$ is unbounded in every neighborhood of the origin even though $f_x(0,0)=0$ exists; differentiability at a point does not require the partial derivatives to be bounded nearby. Combining continuity and the derivative computation: \[\boxed{f\ \text{is continuous and differentiable at } (0,0)}\]