Step 1: Plan before differentiating.
$y$ is a $3\times3$ determinant in $x$. Differentiating a messy determinant directly is hard, so first make it simpler using a column operation, then differentiate.
Step 2: Spot equal columns.
Look at column 1 and column 3 of the top two rows: both have $\sin x$ then $\cos x$. So subtracting column 1 from column 3 will create zeros there. Apply $C_3\to C_3-C_1$.
Step 3: New third column.
Top entry: $\sin x-\sin x=0$. Middle: $\cos x-\cos x=0$. Bottom: $1-x$. So the third column is $\begin{bmatrix}0\\0\\1-x\end{bmatrix}$.
Step 4: Expand along that column.
Only the bottom entry survives, so \[ y=(1-x)\begin{vmatrix}\sin x&\cos x\\ \cos x&-\sin x\end{vmatrix}. \]
Step 5: Evaluate the small determinant.
It equals $(\sin x)(-\sin x)-(\cos x)(\cos x)=-\sin^2x-\cos^2x=-1$. So \[ y=(1-x)(-1)=x-1. \]
Step 6: Differentiate.
Now it is easy: $\dfrac{dy}{dx}=\dfrac{d}{dx}(x-1)=1.$ \[ \boxed{1} \]