I'm trying to find the total derivative $Df(x,y)$ of $f(x,y)=y\cos x$.
I'm familiar with this definition mostly: https://en.wikipedia.org/wiki/Total_derivative#The_total_derivative_as_a_linear_map
I believe there is a theorem that states that if the partial derivatives exist and are continuous, then $f$ is differentiable. I haven't learned this theorem in class however, and would like to avoid using it if possible. Is there any easy ways to see if this derivative even exists on $\mathbb R^2$?
Well, it's unclear exactly what you're allowed to use here, but if a function $f:\mathbb{R}^n\to \mathbb{R}^m$ is $\mathscr{C}^1$ we can write its total derivative in terms of the $m\times n$ Jacobian matrix: $$ \text{Jac}(f)= \begin{bmatrix} \frac{\partial f_1}{\partial x_1}&\cdots&\frac{\partial f_1}{\partial x_n}\\ \vdots&\ddots&\vdots\\ \frac{\partial f_m}{\partial x_n}&\cdots&\frac{\partial f_m}{\partial x_n} \end{bmatrix}. $$ So, for us this is simple because our map is $f:\mathbb{R}^2\to \mathbb{R}$ and we just have to produce a $1\times 2$ matrix of partials. These are $\frac{\partial f}{\partial x}=-y\sin x$ and $\frac{\partial f}{\partial y}=\cos x$. Then the total derivative at $(x_0,y_0)\in \mathbb{R}^2$ is represented by the matrix $$ \text{Jac}(f)_{(x_0,y_0)}= \begin{bmatrix} -y_0\sin x_0 &\cos x_0 \end{bmatrix}.$$