Find the absolute maximum value of $f(x)=2 x^{4} \ln \left(\frac{1}{x}\right)$ and say where it is assumed.

Step 1 ：First, we need to find the derivative of the function. Using the product rule and chain rule, we get \(f'(x) = 2x^{4} \cdot \frac{-1}{x} + 4x^{3} \ln \left(\frac{1}{x}\right) = -2x^{3} + 4x^{3} \ln \left(\frac{1}{x}\right)\).

Step 2 ：Next, we set the derivative equal to zero and solve for \(x\). This gives us \(-2x^{3} + 4x^{3} \ln \left(\frac{1}{x}\right) = 0\). Simplifying, we get \(2x^{3} \ln \left(\frac{1}{x}\right) = 2x^{3}\).

Step 3 ：Dividing both sides by \(2x^{3}\), we get \(\ln \left(\frac{1}{x}\right) = 1\). Exponentiating both sides, we get \(\frac{1}{x} = e\), so \(x = \frac{1}{e}\).

Step 4 ：We also need to check the endpoints of the domain of \(f(x)\), which are \(0\) and \(\infty\). However, as \(x\) approaches \(0\) or \(\infty\), \(f(x)\) approaches \(-\infty\).

Step 5 ：So, the absolute maximum value of \(f(x)\) is \(f\left(\frac{1}{e}\right) = 2\left(\frac{1}{e}\right)^{4} \ln \left(e\right) = \frac{2}{e^{4}}\).

Step 6 ：Finally, we check that this value is indeed a maximum by taking the second derivative of \(f(x)\), \(f''(x) = -6x^{2} + 12x^{2} \ln \left(\frac{1}{x}\right) - 4x^{3} \cdot \frac{1}{x}\).

Step 7 ：Substituting \(x = \frac{1}{e}\) into \(f''(x)\), we get \(f''\left(\frac{1}{e}\right) = -6\left(\frac{1}{e}\right)^{2} + 12\left(\frac{1}{e}\right)^{2} \ln \left(e\right) - 4\left(\frac{1}{e}\right)^{3} \cdot e = -\frac{6}{e^{2}} + \frac{12}{e^{2}} - \frac{4}{e^{2}} < 0\), so \(f\left(\frac{1}{e}\right)\) is indeed a maximum.

Step 8 ：Thus, the absolute maximum value of \(f(x)\) is \(\boxed{\frac{2}{e^{4}}}\), and it is assumed at \(x = \frac{1}{e}\).