Step 1: A handy bound.
For $x\neq 0$, $f(x,y)=\frac{x^2 y}{1+x^2}\sin\frac1x$. Since $|\sin\frac1x|\le 1$ and $\frac{x^2}{1+x^2}\le x^2$, we get $|f(x,y)|\le x^2|y|$.
Step 2: Option A, the repeated limit.
Fix $y$ and let $x\to 0$. By the bound, $f\to 0$. Then letting $y\to 0$ keeps it $0$. So the repeated limit exists. Option A is true.
Step 3: Option C, the $y$ partial.
Since $f$ is linear in $y$, $\frac{\partial f}{\partial y}=\frac{x^2}{1+x^2}\sin\frac1x$ for $x\neq 0$, and $0$ at $x=0$. As $(x,y)\to(0,1)$ this is bounded by $x^2\to 0$, matching the value $0$. So this partial is continuous at $(0,1)$. Option C is true.
Step 4: Option D, the scaled limit.
Look at $\frac{f(x,y)}{\sqrt{x^2+y^2}}$. Using the bound, $\frac{|f|}{\sqrt{x^2+y^2}}\le \frac{x^2|y|}{\sqrt{x^2+y^2}}\le x^2\to 0$. So this limit exists and equals $0$. The claim that it does not exist is wrong, so option D is false.
Step 5: Option B, the $x$ partial.
The factor $\sin\frac1x$ swings wildly as $x\to 0$, so $\frac{\partial f}{\partial x}$ does not settle to a single value. It is not continuous at $(0,1)$. Option B is false.
Step 6: Conclusion.
The true statements are A and C.
\[ \boxed{A,\ C} \]