Clear[x,y];
ContourPlot[x^2*y^3 == (x^2+y^2-1)^3,{x,-2,2},{y,-2,2}]

++++++++++++++++++

The next cell will graph where the function is zero
If your equation doesn' t have zero on one side, then you' ll have to rearrange it.

Clear[x,y,function];
function[x_,y_]=x^2*y^3- (x^2+y^2-1)^3
ContourPlot[function[x,y]==0,{x,-2,2},{y,-2,2}]

?ContourPlot

+++++ homework checking (textbook problems, similar to WileyPLUS) +++++++

(* problem 4 *)
Clear[x,y,function];
function[x_,y_]=x^2*y-2*y+5;
Print[function[x,y]==0]
 ContourPlot[function[x,y]==0,{x,-5,5},{y,-5,5}]

(* problem 5 *)
Clear[x,y,function];
function[x_,y_]=Sqrt[x]-5*Sqrt[y];
Print[function[x,y]==0]
 ContourPlot[function[x,y]==0,{x,0,20},{y,0,20}]

(* problem 11 *)
Clear[x,y,function];
function[x_,y_]=x*Log[y]+y^3-Log[x];
Print[function[x,y]==0]
 ContourPlot[function[x,y]==0,{x,0,1},{y,0,1}]

Clear[y];
FindRoot[function[.2,y]==0,{y,.001}]

Clear[y];
FindRoot[function[.1,y]==0,{y,.001}]

(* problem 20 *)
Clear[x,y,function];
function[x_,y_]=Sin[x*y]-x;
Print[function[x,y]==0]
 ContourPlot[function[x,y]==0,{x,0,2},{y,0,Pi}]

(* problem 23 *)
Clear[x,y,function];
function[x_,y_]=x*y^2-1;
Print[function[x,y]==0]
 ContourPlot[function[x,y]==0,{x,0,2},{y,-2,0}]

(* problem 25 *)
Clear[x,y,function];
function[x_,y_]=x^2/(x*y-4)-y^2;
Print[function[x,y]==0]
 ContourPlot[function[x,y]==0,{x,3,5},{y,1,3}]

Clear[y];
Solve[function[4,y]==0,y];
N[%]

(* problem 28 *)
Clear[x,y,function];
function[x_,y_]=x^2+y^2-4*x+7*y-15;
Print[function[x,y]==0]
 ContourPlot[function[x,y]==0,{x,-10,10},{y,-10,10}]

(* problem 29 *)
Clear[x,y,function];
function[x_,y_]=x^2/25+y^2/9-1;
Print[function[x,y]==0]
 ContourPlot[function[x,y]==0,{x,-5,5},{y,-5,5}]

(* problem 32 *)
(* As an example, I've graphed the original function y=x^2 and
the tangent line y=2x-1, and the circle (x-8)^2+(y-0)^2=45 *)
Clear[x,y,function];
function[x_,y_]= x^2-y;
linefunction[x_,y_]=2x-1-y;
circlefunction[x_,y_]=(x-8)^2+(y-0)^2-45;
Print[function[x,y]==0,"   ",linefunction[x,y]==0,"    ",circlefunction[x,y]==0]
 ContourPlot[{function[x,y]==0,linefunction[x,y]==0,circlefunction[x,y]==0},{x,0,15},{y,0,15}]

++++++++++++ check problem 56 ++++++++
For the writeup, do the calculation yourself, showing all steps clearly.

Clear[a,b,n,R,P,V];
D[(P+((a (n^2))/((V[P])^2))) (-b n+V[P]),P]

Simplify[%]

+++++++++++++OLD STUFF++++++++++++++++++++

Clear[eq,x,y,xmin,xmax,ymin,ymax];
eq=(x^2+x*y-y^3==x*y^2);
xmin=-5;
xmax=5;
ImplicitPlot[eq,{x,xmin,xmax},AspectRatio->Automatic,GridLines->Automatic,PlotStyle->Thickness[0.01]];

Clear[eq,x,y,xmin,xmax,ymin,ymax];
eq=(x^2+x*y[x]-y[x]^3==x*y[x]^2);
D[eq,x]
Solve[%,y'[x]]

ImplicitPlot[E^(x^2)+Log[y]==0,{x,-1,1},GridLines->Automatic,PlotStyle->Thickness[0.01],AspectRatio->1];

Plot[(E^(-(E^(x^2)))),{x,-1,1}]

ContourPlot[ArcTan[(x^2) y]==x (y^2),{x,-0.1`,0.1`},{y,-0.1`,0.1`},GridLines->Automatic,ContourStyle->Thickness[0.01`]]

ContourPlot[ArcTan[(x^2) y]==x (y^2),{x,0.1`,5},{y,0.1`,5},GridLines->Automatic,ContourStyle->Thickness[0.01`]];

Clear[eq,x,y,xmin,xmax,ymin,ymax];
eq=ArcTan[(x^2) y]==x (y^2);
xmin=0.1`;
xmax=5;
ymin=0.1`;
ymax=5;
ContourPlot[eq,{x,xmin,xmax},{y,ymin,ymax},AspectRatio->Automatic,GridLines->Automatic,ContourStyle->Thickness[0.01`]]

Clear[x,y];
ContourPlot[x Log[y]+(y^3)==Log[x],{x,0.1`,5},{y,0.1`,5},AspectRatio->Automatic,GridLines->Automatic,ContourStyle->Thickness[0.01`]];

ContourPlot[Log[x y]==2 x,{x,-1,-0.001`},{y,-60,-0.001`},ContourStyle->Thickness[0.01`],AspectRatio->1,AxesOrigin->{0,0}];

ContourPlot[Log[x y]==2 x,{x,0.001`,1},{y,0.001`,60},ContourStyle->Thickness[0.01`],AspectRatio->1];

Clear[x,y];
ContourPlot[(x^3)+(y^3)-x (y^2)==5,{x,0,2},{y,1,3},AspectRatio->Automatic,GridLines->Automatic,ContourStyle->Thickness[0.01`]];

ImplicitPlot[(x^2)^(1/3)+(y^2)^(1/3)==3^(2/3),{x,-3,3},{y,-3,3}GridLines->Automatic,PlotStyle->Thickness[0.01]];