Perhaps you can do the drawing. Let $L$ be the location of the light. (It is on the ground!) Let $F$ be the location of the man's feet, and let $H$ be the top of his head. Let $B$ be the bottom of the building. Draw the horizontal line $LB$. Draw the vertical line $FH$ to represent the short thin man.
Draw the line through $L$ and $H$. Let it meet the wall of the building at $S$. Then $BS$ is the length of the shadow.
It is convenient to call the distance $LF$ by the name $x$. We are told the man is walking at $3$ feet per second, so $\frac{dx}{dt}=3$.
Let $y=BS$. Note that triangles $LBS$ and $LFH$ are _similar_ , so $$\frac{y}{30}=\frac{4}{x}.$$ We may want to rewrite this as $$y=\frac{120}{x}.$$ Differentiate. We get $$\frac{dy}{dt}=-\frac{120}{x^2}\frac{dx}{dt}.$$ Now it is almost over.