Answer
False
Work Step by Step
As per the definition of a gradient vector
$∇f(x,y)=f_{x}i+f_{y}j$
Therefore, $∇f(x,y)=∇(lny)=\frac{∂(lny)}{∂x}i+\frac{∂(lny)}{∂y}j$
$=\frac{1}{y}j$
Thus, $∇f(x,y)=\frac{1}{y}j$
Hence, $∇f(x,y) \ne \frac{1}{y}$
