Suppose $y$ varies inversely with $x$, and $y=39$ when $x=\frac{1}{3}$. What is the value of $y$ when $x=26$ ?

Step 1 ：The problem states that \(y\) varies inversely with \(x\). This means that the product of \(y\) and \(x\) is a constant. We can express this relationship as \(y = \frac{k}{x}\), where \(k\) is the constant of variation.

Step 2 ：We can find the value of \(k\) using the given values of \(y\) and \(x\). Given that \(y=39\) when \(x=\frac{1}{3}\), we substitute these values into the equation to get \(k = y \times x = 39 \times \frac{1}{3} = 13\).

Step 3 ：Now we can use the value of \(k\) to find \(y\) when \(x=26\). Substituting these values into the equation, we get \(y = \frac{k}{x} = \frac{13}{26} = 0.5\).

Step 4 ：Final Answer: The value of \(y\) when \(x=26\) is \(\boxed{0.5}\).