[tex]\bf e^x-x^3+y^2=10\\\\
-----------------------------\\\\
e^x-3x^2++2y\cfrac{dy}{dx}=0\implies \cfrac{dy}{dx}=\cfrac{3x^2-e^x}{2y}[/tex]
now, that's just the slope of that equation, now, what's the slope of a normal of it?
well, if you recall, a normal is just a perpendicular line to the tangent line at that same point, so the one above is the slope of a tangent line, the slope of a normal to a tangent, is the "negative reciprocal" of that
that is [tex]\bf \textit{perpendicular, negative-reciprocal slope}
\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-----------------------------\\\\
\textit{thus, the slope of the normal is }\left. -\cfrac{2y}{3x^2-e^x}
\right|_{0,3}\implies -\cfrac{2(3)}{3(0)^2-e^0}[/tex]
