How to find Stationary points when there are 2 varaibles.

So question is find stationary points of the curve : x^3 + y^3 -3xy = 48 and determine their nature.
I am not sure. I have tried diffentiating x and then y then rearranging/simultaneuos equations but i keep on getting zeros when the the stationary point is meant to be 2,4 for max and -6^1/3, 6^2/3 for min.

You need to differentiate implicitly with respect to x.
$x^3$ differentiates normally
$y^3$ you differentiate with respect to y, then multiply by $\frac{dy}{dx}$
$-3xy$ you use product rule(-3x and y), when you differentiate y, differentiate it with respect to y then multiply by $\frac{dy}{dx}$
$48$ becomes 0.

Rearrange the final equation to get $\frac{dy}{dx}$ on one side, then find the values of x and y for which the differential equals 0.

Okay so for x^3, i get 3x^2
y^3 is y^3 times by dy/dx
and for -3xy i get -3y+(-3x times by dy/dx)

Just want to make sure if it is right.

The second one is incorrect.

You should have $3x^2 + 3y^{2}\frac{dy}{dx} - 3y - 3x\frac{dy}{dx} = 0$

Ahh okay thanks!. Also i was looking at your stats just found it really weird how in c4 you got 100% but in c1 c2 you got worse ( silly mistakes i guess )

C1, C2 and S1 were mostly self taught in year 11. I didn't have so much time and didn't know the most efficient ways of studying etc. yet, so didn't do so well. You might notice that my worst module last year was equal to my best module the year before.

So question is find stationary points of the curve : x^3 + y^3 -3xy = 48 and determine their nature.
I am not sure. I have tried diffentiating x and then y then rearranging/simultaneuos equations but i keep on getting zeros when the the stationary point is meant to be 2,4 for max and -6^1/3, 6^2/3 for min.