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Let$f\left(x\right)=-x\log \left(3x\right)$ 

1st iteration :

$$\begin{gathered} x0=0.3 and x1=0.4 \\ f\left(x0\right)=f(0.3)=0.0137272 and f\left(x1\right)=f(0.4)=-0.0316725 \\ \therefore x2=x0-f\left(x0\right)\cdot \frac{x1-x0}{f\left(x1\right)-f\left(x0\right) } \\ x2=0.3-0.0137272\cdot \frac{0.4-0.3}{-0.0316725-0.0137272 } \\ x2=0.3302364 \\ \therefore f\left(x2\right)=f(0.3302364)=-0.3302364\log (0.9907092)=0.0013387 \end{gathered}$$


2nd iteration :

$$\begin{gathered} x1=0.4 and x2=0.3302364 \\ f\left(x1\right)=f(0.4)=-0.0316725 and f\left(x2\right)=f(0.3302364)=0.0013387 \\ \therefore x3=x1-f\left(x1\right)\cdot \frac{x2-x1}{f\left(x2\right)-f\left(x1\right) } \\ x3=0.4-(-0.0316725)\cdot \frac{0.3302364-0.4 }{0.0013387-(-0.0316725)} \\ x3=0.3330655 \\ \therefore f\left(x3\right)=f(0.3330655)=-0.3330655\log (0.9991965)=0.0001163 \end{gathered}$$


3rd iteration :

$$\begin{gathered} x2=0.3302364 and x3=0.3330655 \\ f\left(x2\right)=f(0.3302364)=0.0013387 and f\left(x3\right)=f(0.3330655)=0.0001163 \\ \therefore x4=x2-f\left(x2\right)\cdot \frac{x3-x2 }{f\left(x3\right)-f\left(x2\right) } \\ x4=0.3302364-0.0013387\cdot \frac{0.3330655-0.3302364}{0.0001163-0.0013387} \\ x4=0.3333346 \\ \therefore f\left(x4\right)=f(0.3333346)=-0.3333346\log (1.0000038)=-0.0000005 \end{gathered}$$


4th iteration :

$$\begin{gathered} x3=0.3330655 and x4=0.3333346 \\ f\left(x3\right)=f(0.3330655)=0.0001163 and f\left(x4\right)=f(0.3333346)=-0.0000005 \\ \therefore x5=x3-f\left(x3\right)\cdot \frac{x4-x3}{f\left(x4\right)-f\left(x3\right) } \\ x5=0.3330655-0.0001163\cdot \frac{0.3333346-0.3330655 }{-0.0000005-0.0001163 } \\ x5=0.3333333 \\ \therefore f\left(x5\right)=f(0.3333333)=-0.3333333\log (0.9999999)=0 \end{gathered}$$


The approximate root of the equation $-x\log \left(3x\right)=0$ using the Secant method is 0.3333333