параметрическая
x = t^(1/2)
y = t^(1/5)
моё решение
y't = 1/5 * t^(-4/5)
x't = 1/2 * t^(-1/2)
y'x = 2/5 * t^(-4)
y"x = (-16/5) * t^(-5.5)

y = cos(x) * e^(-x)
y' = e^(-x) * cos(x) - e^(-x) * sin(x)
y" = e^(-x) * cos(x) - e^(-x) * sin(x) + e^(-x) * cos(x) - e^(-x) * sin(x)

так ли это?

параметрическая
x = t^(1/2)
y = t^(1/5)
y't = (1/5) * t^(-4/5)
x't = 1/2 * t^(-1/2)
y'x = (y't)/(x't) = (1/5 * t^(-4/5))/(1/2 * t^(-1/2)) = 2/5 * t^(-4/5 + 1/2) = 2/5 * t^(-3/10)
y''xx = (2/5 * (-3/10) * t^(-13/10)) / (1/2 * t^(-1/2)) = -6/25 * t^(-13/10 + 1/2) =
= -6/25 * t^(-4/5)

y = cos x * e^(-x)
y' = -e^(-x) * cos x - e^(-x) * sin x
y" = e^(-x) * cos x + e^(-x) * sin x + e^(-x) * sin x - e^(-x) * cos x = 2 * e^(-x) * sin x