I am trying to calculate the area $y = x^2$ from $0 \leq x \leq 1$, but I get the result is $1/3$. When I try to use polar then I get $r = \tan\left({\theta}\right)\cos\left({\theta}\right)$ from $0\leq \theta \leq \dfrac{\pi}{4}$.
When I use $$A = \dfrac{1}{2} \int_{a}^{b} (r)^2\:\mathrm{d}x = \dfrac{1}{2} \int_{0}^{\frac{\pi}{4}} (\tan\left({\theta}\right)\cos\left({\theta}\right))^{2}\:\mathrm{d}x,$$ the result is $1/6$? Or is there any concept that I at fault with?
You've made multiple errors.
The conversion from cartesian to polar was not done correctly. $y = r\sin\theta, x = r\cos\theta$, so you should have ended up with $r = \sec \theta \tan\theta$ rather than what you got.
Next up, your variable of integration after conversion should be $\theta$ not $x$.
Finally, when you work out that (incorrect) definite integral that you wrote, you do not get $\frac 16$.
When you evaluate the correct integral $\displaystyle \int_0^{\frac{\pi}{4}} \frac 12 r^2 d\theta = \int_0^{\frac{\pi}{4}} \frac 12 \sec^2 \theta\tan^2 \theta$, you get the value $\frac 16$. The indefinite integral is $\frac 12 \cdot \frac 13 \tan^3\theta+c$, which evaluates to $\frac 16$ with those bounds.
This is in fact expected. If you did a sketch of exactly what area you were computing with each method, you'd realise that what you're getting with polar coordinates is the green area between the curve $y=x^2$ and $y=x$, but what the integration with cartesian coordinates gives is the yellow area. As expected the areas add up to the area of the right triangle bounded by vertices $(0,0); (1,0);(1,1)$, i.e. $\frac 13 + \frac 16 = \frac 12$.
Lesson to learn: always sketch the curve before you attempt the area computation.