$y^2+2y=x^3+7x$
at all its intersection points with the line $x=1$
.

Can you differentiate both sides with respect to
$x.$

How can you rearrange your equation to find
$\frac{dy}{dx}?$

Try to substitute
$x=1$
into the original equation.

By differentiating both sides with respect to
$x,$
we have$2y\frac{dy}{dx}+2\frac{dy}{dx}=3x^2+7$
(using the chain rule). Rearranging gives$\frac{dy}{dx}=\frac{3x^2+7}{2y+2}.$
Now we need to find all intersection points with the line$x=1.$
These are given by$y^2+2y8=0,$
which has solutions$y\in\{4,2\}.$
We then plug this into the derivative expression to get the gradient values$\frac{5}{3}$
and$\frac{5}{3}.$