Find $\frac{dy}{dx}$ if $2y^3t+t^3y=1$ and $\frac{dt}{dx}=\frac{1}{\cos t}$ I know it's a parametric equation, where I've to apply chain rule:
$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}$ 
Now, half of the work is done- $\frac{dt}{dx}=\frac{1}{\cos t}$
I differentiated $\frac{dy}{dt}(2y^3t+t^3y)=\frac{dy}{dt}(1)$
$2y^3+6ty^2\frac{dy}{dt}+3t^2y+t^3\frac{dy}{dt}=0$
$\frac{dy}{dt}=\frac{-2y^3-3t^2y}{6y^2t+t^3}$
And combining the above terms, we get-
$\frac{dy}{dx}=\frac{-2y^3-3t^2y}{(6y^2t+t^3){\cos t}}$
I feel I'm doing something wrong, my approach is incorrect because the parametric equations that I've seen contains only $t$ or other parameter after differentiation, not $x$ or $y$.
EDIT- I tried to find the anti-derivative $\frac{1}{\cos t}$, and then plugged that value of $t$ to $2y^3t+t^3y=1$ which made things even more difficult.

Begin by differentiating w.r.t. $x$ on both sides of the equation because you ultimately want to find the derivative of $y$ w.r.t. $x$.
$\frac{d}{dx}(2y^3t+t^3y)=\frac{d}{dx}(1)$
which gives
$\frac{d}{dx}([2y^3]t) +\frac{d}{dx}(t^3y)=0$. 
Now, use the product rule. If it makes it easier, process each term on the left hand side of the equation separately. Processing $\frac{d}{dx}([2y^3]t)$ first:
$\frac{d}{dx}([2y^3]t) = 2y^3 \cdotp \frac{dt}{dx}+ t(2\cdotp3y^2\frac{dy}{dx})$.
Next, process $\frac{d}{dx}(t^3y)$:
$\frac{d}{dx}(t^3y) = t^3\cdotp \frac{dy}{dx} + y \cdotp 3t^2\cdotp \frac{dt}{dx}$.
Now, you just need to plug in the value for $\frac{dt}{dx}$ and solve for $\frac{dy}{dx}$:
$2y^3 \cdotp \frac{dt}{dx}+ t(2\cdotp3y^2\frac{dy}{dx}) + t^3\cdotp \frac{dy}{dx} + y \cdotp 3t^2\cdotp \frac{dt}{dx} = 0$
$2y^3 \cdotp \frac{1}{\cos t}+ t(6y^2\frac{dy}{dx}) + t^3\cdotp \frac{dy}{dx} + y \cdotp 3t^2\cdotp \frac{1}{\cos t} = 0$.
Do you see why we had differentiated w.r.t. $x$ in the beginning? It was because we wanted the $\frac{dy}{dx}$ term eventually. Can you finish the rest of the solution?
