(a) Differentiate implicitly, using the product rule on each term: \[e^{2x}y^{\prime}+2ye^{2x}-2yy^{\prime}e^x-y^2e^x=0.\] Collect y′ and divide by the nonzero exponential: \[(e^x-2y)y^{\prime}=y^2-2ye^x.\] Hence \[\boxed{\frac{dy}{dx}=\frac{2ye^x-y^2}{2y-e^x}}.\]

(b) A vertical tangent requires the denominator to vanish: e^x=2y. Substituting into the original curve gives \[y(2y)^2-y^2(2y)=2,\qquad2y^3=2.\] Thus y=1 and e^x=2, giving \[\boxed{(\ln2,1)}.\] The numerator is then 4−1=3≠0, confirming a vertical tangent.