Slope of the indifference curve

So I am taking mathematical economics and in the HW my professor asked to draw a couple of level curves for $f(x,y)=xy$.

Attempt: So I did it the following way. To find the slope I took the partial derivative with respect to $x$ and then with respect to $y$.:

$$\frac{\partial f}{\partial x}=y\\\frac{\partial f}{\partial y}=x$$

Now to find $\frac{\partial y}{\partial x}$ I just divide the upper equality $(\frac{\partial f}{\partial x})$ by the lower equality ($\frac{\partial f}{\partial y}$):

$$\frac{\partial f}{\partial x} \frac{\partial y}{\partial f}=\frac{\partial y}{\partial x}=\frac{y}{x}$$

So why is it wrong and why should I use the total differential to find the correct slope? Any hints please.

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I don't know why you are focusing somewhat indirectly on finding the slopes of these curves, when you can compute their equations explicitly. Level curves of $f(x,y)$ are curves implicitly defined by equations of the form $f(x,y) = c$ for $c$ constant. So you could draw, for example, graphs of the curves $xy = 1$, $xy=-1$, $xy = 5$, etc. (When $c \neq 0$ the level curve for the value $c$ is a hyperbola with equation $y = c/x$. When $c = 0$ the level curve is the union of the two axes.) –  leslie townes May 26 '12 at 23:14
yeah I know that. But, how can I find the slope of the level curves? Namely, $\frac{\partial y}{\partial x}$. It is a part of the assignment. –  Koba May 26 '12 at 23:17

Level curves have $f(x,y)=C$ for some constant $C$. So you can take: $$C=xy$$ $$y=\frac Cx$$ and then differentiate to get $$\frac{dy}{dx}=-\frac{C}{x^2}$$
This is the slope of the contour plot. You have $z=f(x,y)$. $\frac {\partial z}{\partial x}$ is in the xz plane, $\frac {\partial z}{\partial y}$ is in the yz plane. A level set is in the xy plane, so to get the slope along a level curve you want $\frac {dy}{dx}$. We're dealing with partial derivatives here: $\frac {\partial z}{\partial x} \neq \frac {dz}{dx}$, so the chain rule doesn't work the way you used it. If you want to sketch contour plots i'd just sketch $y=C/x$ for evenly-spaced values of C directly. I don't see why you'd want/need the derivative. –  Robert Mastragostino May 27 '12 at 16:19