Show that there exists the maximum value of the function $f(x,\ y)=(3xy+1)e^{-(x^2+y^2)}$ on $\mathbb{R}^2$, then find the value.
1. **Existence of Maximum:**
   - The function \( f(x, y) = (3xy + 1)e^{-(x^2 + y^2)} \) is continuous on \(\mathbb{R}^2\).
   - At the origin, \( f(0, 0) = (3 \cdot 0 \cdot 0 + 1)e^{-(0^2 + 0^2)} = 1 \), which is positive.
   - As \(|(x, y)| \to \infty\), \( e^{-(x^2 + y^2)} \to 0 \), hence \( f(x, y) \to 0 \).
   - By the Extreme Value Theorem, since \( f \) is continuous and tends to zero at infinity, there must exist a maximum value on \(\mathbb{R}^2\).

2. **Reduction to One Dimension:**
   - For a fixed \( r = \sqrt{x^2 + y^2} \), we can express \( f \) in terms of \( r \).
   - The maximum value of \( 3xy \) for \( x^2 + y^2 = r^2 \) is given by:
     \[
     3xy = \frac{3}{2}(x^2 + y^2) - \frac{3}{2}(x - y)^2 \leq \frac{3}{2}r^2
     \]
   - Therefore, we need to maximize \( f(r) = \left( \frac{3}{2}r^2 + 1 \right)e^{-r^2} \).

3. **Finding the Maximum:**
   - Define \( g(s) = \left( \frac{3}{2}s + 1 \right)e^{-s} \) where \( s = r^2 \).
   - Differentiate \( g(s) \) with respect to \( s \):
     \[
     g'(s) = \left( \frac{3}{2}e^{-s} \right) + \left( \frac{3}{2}s + 1 \right)(-e^{-s}) = \frac{3}{2}e^{-s} - \left( \frac{3}{2}s + 1 \right)e^{-s}
     \]
     \[
     g'(s) = e^{-s} \left( \frac{3}{2} - \left( \frac{3}{2}s + 1 \right) \right) = e^{-s} \left( \frac{3}{2} - \frac{3}{2}s - 1 \right) = e^{-s} \left( \frac{1}{2} - \frac{3}{2}s \right)
     \]
   - Set \( g'(s) = 0 \):
     \[
     e^{-s} \left( \frac{1}{2} - \frac{3}{2}s \right) = 0
     \]
     \[
     \frac{1}{2} - \frac{3}{2}s = 0 \implies s = \frac{1}{3}
     \]
   - Evaluate \( g(s) \) at \( s = \frac{1}{3} \):
     \[
     g\left( \frac{1}{3} \right) = \left( \frac{3}{2} \cdot \frac{1}{3} + 1 \right)e^{-\frac{1}{3}} = \left( \frac{1}{2} + 1 \right)e^{-\frac{1}{3}} = \frac{3}{2}e^{-\frac{1}{3}}
     \]

The final answer is \(\boxed{\frac{3}{2}e^{-\frac{1}{3}}}\)