Firstly you do not need to be given y(5)=-26 You can work that out for yourself.

Like this, sub in x=5 and solve for y

\(5*5^2+5*5+5y=20\\ \text{divide through by 5}\\ 25+5+y=4\\ y=-26\)

So giving you that was probably a bit of a red herring.

\(5x^2+5x+xy=20\\ \text{differentiate both sides}\\ 10x+5+x\frac{dy}{dx}+1y=0\\ \text{Sub in x=5}\\ 50+5+5\frac{dy}{dx}+y=0\\ 5\frac{dy}{dx}=-55-y\\ \text{but we already know that when x=5, y=-26 so sub that in too}\\ 5\frac{dy}{dx}=-55+26\\ 5\frac{dy}{dx}=-29\\ \frac{dy}{dx}=-5.8\\ \)

**y'(5) = -5.8**

LaTex:

5x^2+5x+xy=20\\

\text{differentiate both sides}\\

10x+5+x\frac{dy}{dx}+1y=0\\

\text{Sub in x=5}\\

50+5+5\frac{dy}{dx}+y=0\\

5\frac{dy}{dx}=-55-y\\

\text{but we already know that when x=5, y=-26 so sub that in too}\\

5\frac{dy}{dx}=-55+26\\

5\frac{dy}{dx}=-29\\

\frac{dy}{dx}=-5.8\\