Define $f(x,y)=\frac{xy}{x^2+y^2\ln(x^2)^2}$ if $x\ne0$, and $f(0,y)=0$ if $y\ne0$. Determine whether $\lim_{(x,y)\to(0,0)}f(x,y)$ exists, and find its value is if the limit does exist.
To determine whether \(\lim_{(x,y)\to(0,0)} f(x,y)\) exists, we need to analyze the behavior of the function \(f(x,y)\) as \((x,y)\) approaches \((0,0)\).

1. **Define the function and consider the limit:**
   \[
   f(x,y) = \frac{xy}{x^2 + y^2 \ln(x^2)^2} \quad \text{if} \quad x \ne 0
   \]
   \[
   f(0,y) = 0 \quad \text{if} \quad y \ne 0
   \]
   We need to determine \(\lim_{(x,y) \to (0,0)} f(x,y)\).

2. **Use the inequality \(a^2 + b^2 \ge 2|ab|\):**
   For \((x,y) \ne (0,0)\), we have:
   \[
   x^2 + y^2 \ln(x^2)^2 \ge 2|x y \ln(x^2)|
   \]
   Therefore,
   \[
   |f(x,y)| = \left| \frac{xy}{x^2 + y^2 \ln(x^2)^2} \right| \le \frac{|xy|}{2|xy \ln(x^2)|} = \frac{1}{2|\ln(x^2)|} = \frac{1}{4|\ln|x||}
   \]

3. **Analyze the limit of the upper bound:**
   We need to evaluate:
   \[
   \lim_{x \to 0} \frac{1}{4|\ln|x||}
   \]
   As \(x \to 0\), \(|\ln|x|| \to \infty\). Therefore,
   \[
   \lim_{x \to 0} \frac{1}{4|\ln|x||} = 0
   \]

4. **Conclude the limit:**
   Since \(|f(x,y)| \le \frac{1}{4|\ln|x||}\) and \(\lim_{x \to 0} \frac{1}{4|\ln|x||} = 0\), it follows by the Squeeze Theorem that:
   \[
   \lim_{(x,y) \to (0,0)} f(x,y) = 0
   \]

The final answer is \(\boxed{0}\)