Suppose y$=$f$($x$,$y$)=$xycos$($x$).$

$$f$$y$=$

help $($formulas$)$

Since the function f$($x$,$y$)$ is choosecontinuousnot continuous at the point $(0,0),$ the partial derivative $$f$$y chooseexistsdoes not exist and is choosecontinuousnot continuous at and near the point $(0,0),$ the solution to y$=$f$($x$,$y$)$ chooseexists and is uniquedoes not exist near the point $(0,0).$