The question is:
Find slopes of all lines tangent to the graph of the relation:
y^3 + 15x^15y^2 - 50y = 150 - 150x^5
at the points on its graph where x=1
there are three points for x = 1: $\displaystyle (1,0),~ \left( 1, \frac {-15 + 5 \sqrt{17}}2\right),~\text{and } \left( 1, \frac {-15 - 5 \sqrt{17}}2 \right)$
differentiate implicitly with respect to x and plug each of these points in to find the slopes of each of the three tangent lines.
as a genral solution you can sovle all problems like yours with this approach:
first sort it as:
$\displaystyle f(x,y) = 0$
then use this formula to differntiate:
$\displaystyle \frac {dy}{dx} = -\frac {f^\prime_x}{f^\prime_y}$
it means when you are differntiatig a respect to x then y assumed as constant and in reverse; be sure before differniation that all statements of the equation should be sored on one side and other side should be only Zero(0).