[tex]e^y +19-e^2 = 3x^2 +4y^2\\\\\implies \dfrac{d}{dx}(e^y +19-e^2) = \dfrac{d}{dx}(3x^2 +4y^2)\\\\\implies e^y \dfrac{dy}{dx}+0 = 3\dfrac{d}{dx} x^2 +4\dfrac{d}{dx} y^2\\\\\implies e^y y' = 6x + 8y y'\\\\\implies e^y y' -8yy' = 6x\\\\\implies y'(e^y -8y) = 6x\\\\\implies y' = \dfrac{6x}{e^y -8y}\\\\\text{At point (-1,2)}\\\\y' = \dfrac{6(-1)}{e^2 -8(2)} = - \left(\dfrac{6}{e^2 -16}\right) =-0.25[/tex]
